1) How do I prove the following: Let A = { 6 a + 4...

Explanation:
You prove it by proveing that every number that can be written as $6a+4b$ where a,b are integers can be written as 2m where m is an integer and vice versa.
If $x\in A$ then $x=6a+4b$ for some $a,b\in \mathbb{Z}$.
so..... you prove that there is an $m\in \mathbb{Z}$ so that..... $x=2m$.
Step 2
So $x\in B$. So $A\subset B$.
Then if $y\in B$ then $x=2n$ for some $n\in \mathbb{Z}$.
so .... you prove that there are $a,b\in Z$ so that ... $y=6a+4b$.
So $y\in A$. So $B\subset A$.
So $A\subset B$ and $B\subset A$ so $A=B$.

Step 1
First direction: show that $A\subseteq B$.
Let $x\in A$, then $x=6a+4b$ for some $x\in B$
Since $x=2(3a+2b)$, it is a multiple of two and in particular, $x\in B$.
Step 2
Second direction: show that $B\subseteq A$
Conversely, let $x\in B$, then $x=2a$ for some $a\in \mathbb{Z}$.
Then $x=6a+4(-a)$, so $x\in A$. (You can pick anything in the brackets)
Hence $A=B$