1) $A \setminus B = A \cap \overline B$
2) $(A\cup B)\setminus(A\cap B)=(A \setminus B)\cup(B\setminus A)$
3) $A\cup B=(A \setminus B)\cup (B \setminus A)\cup (A\cap B)$
I need to prove the following equalities of sets. I only have an idea for the first and did the following.
1) My Answer : By definition $A \setminus B = \{x \in U \mid x \in A \text{ and } x\notin B\}$.
For set B we have the following : $$x \in U\text{ and }\space x\in B$$ but, $$\overline B=\{x\in U\mid x\notin B\}$$
Hence from the following facts I got,
$$A\cap \overline B=\{x\in U\mid x\in A \text{ and }\notin B.$$
I assume that this is logically true but I'm not sure whether it is true as a proof.
I'll be waiting for your comments about "1"(to my solution) and also help for "2 and 3"
The usual thing to do in this kind of proof is to show that each set (on each side of the equation) is contained in the other.
As for example in 1) you need to verify that $A\setminus B\subseteq A\cap \overline{B}$ and $A\cap \overline{B}\subseteq A\setminus B$. And this you can do taking an $x$ in $A\setminus B$ and proving that $x\in A\cap \overline{B}$ and vice-versa. But the idea to prove this is pretty similar to what you've done.
That being said, the statement 2) is false, maybe you got it written wrong because if you change the second part for $(A\setminus B)\cap(B\setminus A)$ it works.