(1) To show that two sets are equal, show that each is a subset of the other. Here that means showing that $A\cap B\subseteq A\setminus(A\setminus B)$ and that $A\setminus(A\setminus B)\subseteq A\cap B$. (I use $A\setminus B$ for set difference instead of the rather old-fashioned $A-B$.)
To show that $A\cap B\subseteq A\setminus(A\setminus B)$, let $x$ be an arbitrary element of $A\cap B$, and show that $x\in A\setminus(A\setminus B)$. Since $x\in A\cap B$, we know that $x\in A$ and $x\in B$. In order to show that $x\in A\setminus(A\setminus B)$, we have to show that $x\in A$ and $x\notin A\setminus B$. We know that $x\in A$, so that part’s no problem. And since $x\in B$, $x\notin A\setminus B$, and we’re done: $x\in A$ and $x\notin A\setminus B$, so $x\in A\setminus(A\setminus B)$. We’ve just shown that every element of $A\cap B$ also belongs to $A\setminus(A\setminus B)$, which is exactly what it means to say that $A\cap B\subseteq A\setminus(A\setminus B)$.
To show that $A\setminus(A\setminus B)\subseteq A\cap B$, let $x\in A\setminus(A\setminus B)$ be arbitrary, and show that $x\in A\cap B$. Since $x\in A\setminus(A\setminus B)$, we know that $x\in A$ and $x\notin A\setminus B$. To finish showing that $x\in A\cap B$, we just have to show that $x\in B$. But since $x\in A$ and $x\notin A\setminus B$, it must be the case that $x\in B$: if $x$ weren’t an element of $B$, it would be in $A\setminus B$, and it isn’t. Thus, $x\in A\cap B$, and $A\setminus(A\setminus B)\subseteq A\cap B$.
Putting the two pieces together, we see that $A\cap B=A\setminus(A\setminus B)$.
(2) Approach these in the same way. To show that
$$E\setminus\bigcap_{j=1}^nA_j=\bigcup_{j=1}^n(E\setminus A_j)\;,$$
prove that
$$E\setminus\bigcap_{j=1}^nA_j\subseteq\bigcup_{j=1}^n(E\setminus A_j)\quad\text{and}\quad\bigcup_{j=1}^n(E\setminus A_j)\subseteq E\setminus\bigcap_{j=1}^nA_j\;.$$
I’ll get you started.
Suppose that $x\in E\setminus\bigcap_{j=1}^nA_j$. Then $x\in E$, and $x\notin\bigcap_{j=1}^nA_j$. The latter means that there is at least one $k$ such that $1\le k\le n$ and $x\notin A_k$. But then $$x\in E\setminus A_k\subseteq\bigcup_{j=1}^n(E\setminus A_j)\,,$$ and it follows that $E\setminus\bigcap_{j=1}^nA_j\subseteq\bigcup_{j=1}^n(E\setminus A_j)$.
Now suppose that $x\in\bigcup_{j=1}^n(E\setminus A_j)$. Then there is some $k$, $1\le k\le n$, such that $x\in E\setminus A_k$. In other words, $x\in E$ and $x\notin A_k$. $A_k\supseteq\bigcap_{j=1}^nA_j$, so if $x\notin A_k$, then certainly $x\notin\bigcap_{j=1}^nA_j$, and since $x\in E$, we have $x\in E\setminus\bigcap_{j=1}^nA_j$. It follows that $\bigcup_{j=1}^n(E\setminus A_j)\subseteq E\setminus\bigcap_{j=1}^nA_j$.
Putting the pieces together yields the first of the two De Morgan’s laws. I’ll leave the second one for you; you should take the same basic approach, though of course the details will be different.
(3) And you want the same approach here: show that $A\times B\subseteq(A\times B_1)\cup(A\times B_2)$ and $(A\times B_1)\cup(A\times B_2)\subseteq A\times B$. The second of these is very easy. If $\langle a,b\rangle\in(A\times B_1)\cup(A\times B_2)$, then $\langle a,b\rangle\in A\times B_1$, or $\langle a,b\rangle\in A\times B_2$. Suppose that $\langle a,b\rangle\in A\times B_1$. Then $a\in A$ and $b\in B_1\subseteq B$, so $\langle a,b\rangle\in A\times B$. A similar argument shows that if $\langle a,b\rangle\in A\times B_2$, then $\langle a,b\rangle\in A\times B$, and it follows that $(A\times B_1)\cup(A\times B_2)\subseteq A\times B$.
The other direction is just as straightforward; I’ll leave it to you.
Yes, to prove that $(A\setminus B)\cap(B\setminus A)=\varnothing$ for all $A$ and $B$, you can begin by assuming that $(A\setminus B)\cap(B\setminus A)\ne\varnothing$. Then let $x\in(A\setminus B)\cap(B\setminus A)$; clearly $x\in A\setminus B$, so $x\in A$. But also $x\in B\setminus A$, so $x\notin A$. This is a contradiction, so the original assumption that $(A\setminus B)\cap(B\setminus A)\ne\varnothing$ must be false. If $p$ is this assumption, and $q$ is $x\in A$, then the argument shows that $p\to(q\land\neg q)$. However
$$\Big(p\to(q\land\neg q)\Big)\to\neg p$$
is a tautology, so we infer $\neg p$, i.e., that $(A\setminus B)\cap(B\setminus A)=\varnothing$.
I would prove the second one by showing that
$$|A\cup B|=|A|+|B|-|A\cap B|\;;$$
it follows immediately that if $|A\cup B|=|A|+|B|$, then $|A\cap B|=0$ and hence that $A\cap B=\varnothing$.
Best Answer
They are fine. Note that the way you've written them reduces the set theory property of distributivity to the logical property (i.e. in a Boolean algebra) of distributivity. This you do in the transition from "$x \in A$ and $(x \in B \text { or } x \in C)$" to "$(x \in A \text { and } x \in B) \text { or } (x \in A \text { and } x \in C)$", in the first proof, e.g.
You can also use a more "natural deduction"-like approach: e.g. for the first inclusion: suppose $x \in A \cap (B \cup C)$. This means that $x \in A$ and also $x \in (B \cup C)$. Now we consider two cases:
Suppose $x \in B$, then $x \in A$ still and $x \in B$ so $x \in A \cap B$, and so certainly $x \in (A \cap B) \cup (A \cap C)$.
On the other hand, if $x \in C$ (second case), $x \in A \cap C$ and again $x \in (A \cap B) \cup (A \cap C)$, and so $x$ is in the right hand side in both cases, and this proves the inclusion.