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Libgdx Y axis, how to invert it?

How to invert Y axis? When I touch on bottom or top of the screen, the Y value is opposite I want

You can't invert an axis per se. You see, in computer graphics, the 2D coordinate system is a bit different from the canonical one taught at school in maths. The difference is that in computer graphics the y-axis is in the opposite direction, that is, from the origin to the bottom it has positive values and from the origin to the top it has negative values. Also, the origin is at the top left corner of the screen. If you can't get used to it then you can always take the opposite value to what you get, for this, asume ycoord holds the value obtained then you can do ycoord = -ycoord and that will get you the value as you're used to. Also, if you want the origin to be in the bottom left corner then you should check your y-coordinate, if it's positive then substract the vertical resolution to it, and if it's negative then add the vertical resolution to it.
But keep mind that you're going against the standard definition for coordinate systems in computer graphics.

I would say this is a duplicate questions of this one:
Move a shape to place where my finger is touched
Check on my answer there, so I won't repeat my self.
Or in short - use camera.unproject() method to get world coordinates from screen coordinates.

How to do 2D ground with depth sense in libgdx?

I know do a horizontal and vertical scroller game (like Mario), in this game type, the character is always in the same distance from user. The character only moves to left and right in horizontal scroller and to down and up in vertical scroller.
But there are games in 2D that the character can move freely in the scene, like for example graphic adventures.
So how can I do that ground in order to move the character freely on the ground with depth sense?
An example can see in this video: http://youtu.be/DbZZVbF8fZY?t=4m17s
Thanks you.

This is how I would do that:
First imagine that you are looking at your scene from the top to the ground. Set your coordinate system like that. So all object on your scene will have X and Y coordinates. All your object movements and checking (when character bumps into a wall or something), calculations do in that 2D world.
Now, to draw your world, if you want simpler thing, without some isometric perspective 3D look you just to draw your background image first, and then order all your objects far to near and draw them that way. Devide your Y coords to squeeze movement area a bit. Add some constant to Y to move that area down. If you characters can jump or fly (move trough Y axe) just move Y coord to for some amount.
But if you want it to be more 3D you'll have to make some kind of perspective transformation - multiply your X coordinate with Y and some constant (start with value 1 for constant and tune it up until optimal value). You can do similar thing with Y coord too, but don't think it's needed for adventure games like this.
This is probably hard to understand without the image, but it's actually very simple transformation.

Drawing a square by dragging the mouse

I am writing a Java program that closely mimics Microsoft Paint. It can draw four different shapes: Lines, Ovals, Rectangles, and Squares. I am very close to finishing this but I am stuck on the logic for drawing squares.
There are two Points involved while drawing these shapes. The first Point(point1) is when the user presses the mouse button and the second Point(point2) is while the user drags the mouse across the canvas. I believe drawing Rectangles and Squares should be quite similar but the part that has confused me is when drawing a square the sides are equal length so point2 isn't exactly where the mouse is.
Here is the fillRect() method header for reference:
fillRect(x, y, width, height)
My functioning code for drawing rectangles is as follows:
g.fillRect((((point1.x < point2.x) ? point1.x : point2.x)),
((point1.y < point2.y) ? point1.y : point2.y),
Math.abs(point2.x - point1.x),
Math.abs(point2.y - point1.y));
I tried using the same code for drawing squares except changing the height parameter to be equal to the width parameter because squares have equal length sides:
g.fillRect((((point1.x < point2.x) ? point1.x : point2.x)),
((point1.y < point2.y) ? point1.y : point2.y),
Math.abs(point2.x - point1.x),
Math.abs(point2.x - point1.x)); //same as width
I don't know what the problem is with making both width and height equal. It works when drawing downwards to the left or right but of course the shape does not expand if you pull straight down. Clicking and dragging up does not work; the square simply moves up with the mouse along the Y axis instead of expanding.
Could anyone point me in the right direction regarding the logic for drawing square from two points?
I can explain my code a bit better if need be.

If I were using a tool to draw a square, I would expect that my mouse cursor would remain on one of the sides of the square while drawing. If you always use the x distance as the side, then if I've drawn farther down than across, the cursor will be outside the square.
I think, while the dragging is going on, the code would need to calculate whether the x distance or the y distance to the origin is the longer, and use that as the side of the square-in-progress. Then the cursor will be on one of the sides, and that side will extend out beyond the cursor to the corner of the square-in-progress.
I don't know why the drawing is going awry -- it is difficult to be sure I understand what you say is going wrong.

You should do two calculations in your MouseDragged method -- first you should calculate the two corner Points of the rectangle, p1, and p2, and only then should you calculate the width. Point p1 is easy -- it's always the first Point pressed, but p2 will require a simple calculation. Once these are clarified, your calculations should fall out correctly.
i.e., something like:
int width = Math.abs(pointA.x - pointB.x);
int height = Math.abs(pointA.y - pointB.y);
width = Math.max(width, height);
height = width;
int x = pointA.x > pointB.x ? pointA.x - width : pointA.x;
int y = pointA.y > pointB.y ? pointA.y - width : pointA.y;
g.fillRect(x, y, width, height);

Setting the pivot point of a JPanel to its center

I am building a Java application that is going to feature two circles of random sizes that need to be clicked by the user. The time between the click on the first and the second circle is going to be measured. Unfortunately, since I am new to Java so things have been slow for me. Currently I have my application draw circles and measure time between clicks using System.nanoTime() but now I am running into a problem.
Because the circles need to be a fixed distance away from eachother I want to use the center of the circles as the origin points. So basically I want to be able to provide coordinates for the circle so that the center of the circle should be at those coordinates. The distance between the circles then describes the distance between the centers. My circle currently is embedded into a JPanel but if I set the JPanel's position it moves the top left to that position.
Of course I have done some searching read that I may need to play around with either AffineTransform or Graphics2D.translate() which I have tried in paintComponent() but this got a bit confusing so then I tried to override setlocation and subtract the radius from the position. It sort of works but it is not the most clean solution. Can aonyone give me some pointers on how to do this?
Thanks in advance.

If I understand the problem statement, all such pairs of circles will lie on opposite sides of a circle centered in the enclosing panel, as shown here. Simply choose a random 0 ≤ θ < π and find its opposite at π - θ. Note how the example's rendering scales as the panel is resized.
As an aside, the example uses setPreferredSize() to establish the dimensions of the drawing panel, but you may want to override getPreferredSize() instead.
Addendum: The example uses fillOval() to render the circles, but you can use draw() with any desired Shape; the latter provides several contains() methods suitable for hit testing, as mentioned here.

You have the coordinates for the two center for the circle (x1, y1) and (x2, y2).
The size of the radius is random.
Once you have the radius of the two, r1 and r2, simply position them at (x1-r1, y1-r1) and (x2-r2, y2-r2).
You can use java.awt.Point to represent the center, and use
center.translate(-radius, -radius)
and use the new translated value as position for the drawing.
Maybe you think it is not a clean solution, but why not? Everything in Java is painted by giving the top left corner for the position, so is the use of the center that is not clean :).
To calculate the left top position by doing -radius is clean :)

Making fillRect fill dynamically from right to left

Hi i am currently working on a Java app and i have to draw a little using Canvas, Graphics etc..
So if i click a point and drag it across, it should have a line drawn in between (Think of drawing a line in paint).
I am currently using fillRect and the question is, is there a way to fillRect from right to left? Or do i have to explicitly create a workaround for this?

fillRect() method needs four arguments, x, y, width, height. What you should do is just compute the values of those arguments. If you want to draw fillRect from right to left, you just need to decrease x and increase width perhaps as your mouse move. That's it.

To add to ntalbs excellent answer (+1).
Basically, when the user clicks a point, you need to store that as the anchor point. When they drag the mouse, you need to determine in which direction the mouse has dragged.
If the click.x > drag.x, the the drag.x becomes the x parameter for your rectangle, otherwise it's the click.x.
Width and height are simple determine as the difference between the click and drag points (taking into consideration which is larger ;))