Most of them will become clearer if you draw Venn diagrams to illustrate them.
For (2), note that it’s always true that $B\subseteq A\cup B$, so what you really need to show is that $A\subseteq B$ if and only if $A\cup B\subseteq B$. Certainly if $A\subseteq B$, then $A\cup B\subseteq B$, since it’s trivially true that $B\subseteq B$. On the other hand, $A\subseteq A\cup B$, so if $A\cup B\subseteq B$, then $A\subseteq B$.
For (3), suppose that $A\subseteq B$. If $x\in B^c$, then $x\notin B$, so certainly $x\notin A$; but that means that $x\in A^c$, and we’ve shown that $B^c\subseteq A^c$. Now suppose that $B^c\subseteq A^c$. By what we just proved, we know that $(A^c)^c\subseteq(B^c)^c$. But $(A^c)^c=A$ and $(B^c)^c=B$, so $A\subseteq B$, as desired.
(5) is really very clear if you draw a Venn diagram, but you can argue as follows. Suppose that $A\subseteq B$. Let $x\in U$ be arbitrary. If $x\in B$, then certainly $x\in B\cup A^c$. If $x\notin B$, then $x\notin A$, so $x\in A^c$, and again we see that $x\in B\cup A^c$. Thus, everything in $U$ is in $B\cup A^c$, and since $B\cup A^c\subseteq U$, we must have $B\cup A^c=U$. Conversely, suppose that $B\cup A^c=U$, and let $x\in A$. Then $x\notin A^c$, but certainly $x\in U=B\cup A^c$, so it can only be that $x\in B$. This shows that $A\subseteq B$.