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4

Clearly not. Let $A=\{a\}$ and $B=\{aa\}$. Now, $A\cap B = \emptyset$ so $(A\cap B)^* = \{\epsilon\}$ but $A^*\cap B^*=B^*=\{a^{2i} : i \in \mathbb{N}\}$ (all strings consisting of an even number of $a$).

3

It seems you have hit the reasoning correctly: A set $S$ is a subset of another set $A$ if $w\in S\implies w\in A$. Since you are trying to show that $\forall w \in A^+ \implies w \in A^*$, you are proving that $A^+ \subseteq A^*$. One more legant way to formally do this is by finding a bijection between the two sets. Since $|A^*|$ = $|\Bbb N|$ you need to ...

2

The "$x\in A^{+}$. By definition $x\in A^{+} \wedge x\in A^{*}$ for all $x\neq A^{0}$" is a bit weird. Not even from a computer science point of view but from a set theory or general mathematical point of view. First, you're saying the same thing again with "$x\in A^{+}$" which you shouldn't. Then, you're saying $x\in A^{*}$ which is the desired result. I ...

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