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I want to Capture a alphanumeric group without underscore

I want to Capture an alphanumeric group in regex such that it does not capture starting underscore. For example _reverse(abc) should return reverse(. I am using (?<name>\w+) but it return _reverse(.

You can try this,
[^a-zA-Z0-9()\\s+]
The output will be reverse(abc)

You can specify characters explicitly, e.g.:
[a-zA-Z0-9]+

From what you are showing, I assume you want to strip underscores and content behind the opening parentheses.
Basically, that should work with a regex like this:
"_([a-zA-Z0-9]+\()"
this can be used in conjunction with a Matcher to extract all capturing groups (in this case, [a-zA-Z0-9]+\() and return them.
Note that you can find almost all the help you need with Regular Expressions on utility sites like RegEx 101 and RegEx Per, the latter being a nice visualizer but only working with javaScript-like expressions.
Also, RegEx 101 contains a Regex Debugger to help avoid dangerous regular expressions

Add Dash to Java Regex

I am trying to modify an existing Regex expression being pulled in from a properties file from a Java program that someone else built.
The current Regex expression used to match an email address is -
RR.emailRegex=^[a-zA-Z0-9_\\.]+#[a-zA-Z0-9_]+\\.[a-zA-Z0-9_]+$
That matches email addresses such as abc.xyz#example.com, but now some email addresses have dashes in them such as abc-def.xyz#example.com and those are failing the Regex pattern match.
What would my new Regex expression be to add the dash to that regular expression match or is there a better way to represent that?

Basing on the regex you are using, you can add the dash into your character class:
RR.emailRegex=^[a-zA-Z0-9_\\.]+#[a-zA-Z0-9_]+\\.[a-zA-Z0-9_]+$
add
RR.emailRegex=^[a-zA-Z0-9_\\.-]+#[a-zA-Z0-9_-]+\\.[a-zA-Z0-9_-]+$
Btw, you can shorten your regex like this:
RR.emailRegex=^[\\w.-]+#[\\w-]+\\.[\\w-]+$
Anyway, I would use Apache EmailValidator instead like this:
if (EmailValidator.getInstance().isValid(email)) ....

Meaning of - inside a character class is different than used elsewhere. Inside character class - denotes range. e.g. 0-9. If you want to include -, write it in beginning or ending of character class like [-0-9] or [0-9-].
You also don't need to escape . inside character class because it is treated as . literally inside character class.
Your regex can be simplified further. \w denotes [A-Za-z0-9_]. So you can use
^[-\w.]+#[\w]+\.[\w]+$
In Java, this can be written as
^[-\\w.]+#[\\w]+\\.[\\w]+$

^[a-zA-Z0-9_\\.\\-]+#[a-zA-Z0-9_]+\\.[a-zA-Z0-9_]+$
Should solve your problem. In regex you need to escape anything that has meaning in the Regex engine (eg. -, ?, *, etc.).

The correct Regex fix is below.
OLD Regex Expression
^[a-zA-Z0-9_\\.]+#[a-zA-Z0-9_]+\\.[a-zA-Z0-9_]+$
NEW Regex Expression
^[a-zA-Z0-9_.+-]+#[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+$

Actually I read this post it covers all special cases, so the best one that's work correctly with java is
String pattern ="(?:[a-zA-Z0-9!#$%&'*+/=?^_`{|}~-]+(?:\\.[a-zA-Z0-9!#$%&'*+/=?^_`{|}~-]+)*|\"(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21\\x23-\\x5b\\x5d-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])*\")#(?:(?:[a-zA-Z0-9](?:[a-zA-Z0-9-]*[a-zA-Z0-9])?\\.)+[a-zA-Z0-9](?:[a-zA-Z0-9-]*[a-zA-Z0-9])?|\\[(?:(?:(2(5[0-5]|[0-4][0-9])|1[0-9][0-9]|[1-9]?[0-9]))\\.){3}(?:(2(5[0-5]|[0-4][0-9])|1[0-9][0-9]|[1-9]?[0-9])|[a-zA-Z0-9-]*[a-zA-Z0-9]:(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21-\\x5a\\x53-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])+)\\])";

Translate phone-number regex to java

I found this regex in a plugin called hi5validator for jQuery, and I found it pretty good, I'm already using it on JavaScript:
/^([\+][0-9]{1,3}([ \.\-])?)?([\(][0-9]{1,6}[\)])?([0-9 \.\-]{1,32})(([A-Za-z \:]{1,11})?[0-9]{1,4}?)$/
I wanted to use this regex but in Java, and I tried to do this same thing with another regex in that library, but when I used an online evaluator, the expression gave lots of trouble. Fortunately, i found another regex that helped with that.
As for this one, can someone give me the proper Java version?

The logic of your regex is fine - you need to fix some minor details:
Put double quotes " instead of slashes / around your regex
Do not escape with back slashes parentheses ( ), dashes - in trailing positions, pluses +, colons :, and dots . inside character classes (I am not sure if it is necessary to escape these characters in Javascript either).
Here is what you should get:
"^([+][0-9]{1,3}([ .-])?)?([(][0-9]{1,6}[)])?([0-9 .-]{1,32})(([A-Za-z :]{1,11})?[0-9]{1,4}?)$"

Java Regex doesn't work with special chars

I got a problem with my parser. I want to read an image-link on a webiste and this normally works fine. But today I got a link that contains special chars and the usual regex did not work.
This is how my code looks like.
Pattern t = Pattern.compile(regex.trim());
Matcher x = t.matcher(content[i].toString());
if(x.find())
{
values[i] = x.group(1);
}
And this is the part of html, that causes trouble
<div class="open-zoomview zoomlink" itemscope="" itemtype="http://schema.org/Product">
<img class="zoomLink productImage" src="
http://tnm.scene7.com/is/image/TNM/template_335x300?$plus_335x300$&$image=is{TNM/1098845000_prod_001}&$ausverkauft=1&$0prozent=1&$versandkostenfrei=0" alt="Produkt Atika HB 60 Benzin-Heckenschere" title="Produkt Atika HB 60 Benzin-Heckenschere" itemprop="image" />
</div>
And this is the regex I am using to get the part in the src-attribute:
<img .*src="(.*?)" .*>
I believe that it has something to do with all the special character inside the link. But I'm not sure how to escape all of them. I Already tried
Pattern.quote(content[i].toString())
But the outcome was the same: nothing found.

The . character usually only matches everything except new line characters. Therefore, your pattern won't match if there are newlines in the img-tag.
Use Pattern.compile(..., Pattern.DOTALL) or prepend your pattern with (?s).
In dotall mode, the expression . matches any character, including a
line terminator. By default this expression does not match line
terminators.
http://docs.oracle.com/javase/1.5.0/docs/api/java/util/regex/Pattern.html#DOTALL

You regex should be like:
String regex = "<img .*src=\"(.*?)\" .*>";

This probably caused by the newline within the tag. The . character won't match it.
Did you consider not using regex to parse HTML? Using regex for HTML parsing is notoriously fragile construct. Please consider using a parsing library such as JSoup for this.

You should actually use <img\\s\\.*?\\bsrc=["'](\\.*?)["']\\.*?> with (?s) modifier.

Regex - escape or character block?

What is the best approach if for instance a question mark is expected in a String.
...[?]...
or
...\?...
Example:
The text bla?bla will match both with the pattern bla[?]bla and bla\?bla (bot not bla?bla obviously) but is there any reason to use one over the other?

There is no technical reason to prefer one over the other: They are equivalent expressions. The character class is only used to avoid entering a backslash, so IMHO the escaped version is "cleaner"
However the reason may be to avoid double-escaping the slash on input. In languages like java, the literal version of the escaped version would look like this:
// in java you need to escape a backslash with another backslash :(
String regex = "...\\?...";
It could be that wherever the regexes are coming from has a similar issue and it's easier to read [?] than \\?