A bunch of questions on $\mathbb{Z}_n$ I'm studying the remainder class $\mathbb{Z}_n$, I've grabbed something, but something else is unfocused. Let 
$$20x \equiv 4\pmod{34}$$
then GCD(20,34)=2 so I rewrite as:
$$10x \equiv 2\pmod{17}$$
and successively:
$$10x \equiv 1\pmod{17}$$
Now I know $\gcd(10, 17)=1$

Question 1: Why? Is this cause I've divided both $20$ and $34$ for the
  same $\gcd=2$? If $d$ is a $\gcd$ for $a$ and $b$, then $\gcd(a/d, b/d)=1$?

At this point I can:
$$1=10\alpha + 17\beta$$
that will be:
$$1=10(-5)+17\cdot 3$$

Question 2: I know that $-5$ is the solution of equation, but why I have to choose $-5$ more than $3$?

Now $-5$ is a solution for:
$$10x \equiv 1\pmod{17}$$
and $-5\cdot2$ is a solution for:
$$10x \equiv 2\pmod{17}$$
$[-10]_{17}$ is class, the entire set is $\{-10+17k: k\in\mathbb{Z}\}$.

Question 3: This set will contain all invertible elements. Right? Now
  If I took $[8]_{17}$ and $[7]_{17}$ why the class $[56]_{17}$ is not
  in the same class of $[1]_{17}$ since it is invertible, matter of
  fact, $\gcd(56,17)=1$? Is $\gcd(x,y)=1$ the only way to test if $[x]_y$ is
  invertible? I know that element $a$ was invertible if $ba=ab=1$, but this
  contradicts my knowledge.


It is certainly true that if $\gcd(a,b)=d$ then $\gcd(a/d,b/d)=1$. 
It is hard to say much more about your calculations, since there was an early problem. (And, what is worse, you ended up with the correct answer despite the early error! That is discussed at the end of this post.) 
It is certainly true that $20x\equiv 4\pmod{34}$ if and only if $10x\equiv 2\pmod{17}$.  For a justification, note that $34$ divides $20x-4$ if and only if $17$ divides $10x-2$.  Maybe this is not obvious to you, so we do details.  Note that  $34$ divides $20x-4$ if and only if for some integer $q$, we have $34q=20x-4$. This is the case if and only if $17q=10x-2$. 
Since $2$ and $17$ are relatively prime, $10x\equiv 2\pmod {17}$ if and only if $5x\equiv 1 \pmod {17}$.  That is where things started to go wrong, since you had $10x\equiv 1\pmod{17}$ instead.  
You ask for justification for the step that led you to $10x \equiv 1 \pmod{17}$. One cannot give such a justification, because the step is not correct.
But as to the step that I used to get to $5x\equiv 1\pmod{17}$, this can be done.
For $10x\equiv 2 \pmod{17}$ if and only if $17$ divides $10x-2$, or equivalently if and only if $17$ divides $2(5x-1)$. Since $17$ and $2$ are relatively prime, $17$ divides $2(5x-1)$ if and only if $17$ divides $5x-1$. In congruence language that says $5x\equiv 1\pmod{17}$.
So we are trying to solve the congruence $5x\equiv 1\pmod{17}$. Now that you have the right congruence, you should be able to push the rest through.
For your Question $2$, you say, correctly, that if we are looking for a solution of $1=10\alpha+17\beta$, then $\alpha=-5$ (with $\beta=3$) is a solution. You ask why we have to take $-5$ "more than $3$." I interpret that as meaning why do we use $\alpha$, not $\beta$.  
The answer is that from $1=(10)(-5)+ (17)(3)$ we can conclude that $17$ divides $(10)(-5)-1$, so $(10)(-5)\equiv 1 \pmod {17}$, which was what you wanted. (But it was not what you actually needed.)  We actually don't care about the value of $\beta$, as long as it is an integer. 
Back to solving the problem. We want to solve $5x\equiv 1 \pmod {17}$.  So we try to come up with $\alpha$ and $\beta$ such that $5\alpha+17\beta=1$.
Note that $\alpha=-10$, $\beta=3$ works.  So the solution to our congruence is $x=-10$.  Equivalently, and more simply, we can take $x=7$, since $7\equiv -10\pmod{17}$.
So the integers $x$ that satisfy our original congruence are all integers of the form $7+17k$, where $k$ ranges over all the integers.
An accident: You ended up with the solutions $-10+17k$, which is correct, and equivalent to the answer I arrived at.  But this was because of two errors that cancelled.  You wrote down the wrong congruence, $10x\equiv 1$ instead of $5x\equiv 1$.  You noted the solution $-5$.  Then you wrote that $-5\cdot 2$ is a solution of $10x\equiv 2\pmod {17}$.  So you multiplied the right hand side by $2$, but not the left.  This cancelled the error you had made earlier, when you wrote $10 x\equiv 1 \pmod{17}$ instead of $5x\equiv 1 \pmod {17}$.  
You could deliberately make these cancelling errors part of your standard procedure, and always get the right answer.  But you will lose contact with what is really going on. 
