Proofs involving $ (A\setminus B) \cup (A \cap B)$
(a). see here .
(b). If $x\in A \setminus B,$ then $x\notin B$. Hence $x\notin A\cap B$
Since $A\setminus B\subseteq A$ and $A\cap B\subseteq A$, we have $(A\setminus B)\cup(A\cap B)\subseteq A$.
Let $a\in A$. There are two cases: either $a\in B$ or $a\notin B$.
If $a\in B$, then $a\in A\cap B$; if $a\notin B$, then $a\in A\setminus B$. Hence $A\subseteq(A\setminus B)\cup(A\cap B)$.
Suppose $x\in(A\setminus B)\cap(A\cap B)$. Then $x\in A\setminus B$ implies $x\notin B$. On the other hand…
$x\in A\cap B$ implies $x\in B$.