Suppose $c$ is in $C$. We want to show that $c$ is in $B$. Certainly $C$ is in $A \cup C$, and so by your first assumption, $c$ is in $A \cup B$. That is, either $c$ is in $A$ or $c$ is in $B$. In the latter case we are done. In the former case, $c$ is in $C$ and in $A$ and so $c$ is in $A \cap C$, and so by your second assumption, $c$ is in $A \cap B$ and hence in $B$.
Thus in all cases, if $c$ is in $C$, then $c$ is in $B$, and so we have shown $C \subset B$.
(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.
Best Answer
NO : $x∉A∪B$ means that $x∉A$ and $x∉B$.
To be element of the union of two sets means to belong at least to one of them.
Thus, if $x$ does not belong to the union of $A$ and $B$, it means that $x$ cannot be element neither of $A$ nor of $B$.