Intersections of some context-free languages
Notice that $L_3 = \Sigma^*$, since you can always choose $y=\epsilon$ (the empty string is a substring of any string) and then choose $x$ from $\Sigma^*$. $L_1$ is also $\Sigma^*$, since $L$ is always a subset of $L^*$ and we're basically saying $L_1 = (\Sigma^*)^*$. These observations imply the following: the answers to #2, #3 and #4 are "yes" (indeed, there's a DFA with one state accepting each of these, which are all $\Sigma^*$)... and the intersection in #1 is simply $L_2$, so the question is really "is $L_2$ context-free?"

Please also note that we have other questions on the site about proving and disproving that languages are context-free.

To clarify:


$L_2 \cap L_3 = L_2 \cap \Sigma^* = L_2$
$L_2 \cup L_3 = L_2 \cup \Sigma^* = \Sigma^*$
$L_1 \cap L_3 = \Sigma^* \cap \Sigma^* = \Sigma^*$
$L_1 \cup L_2 = \Sigma^* \cup L_2 = \Sigma^*$


As for $L_2$:

First note that I'm assuming $(\Sigma + \Sigma)^*$ is a funny way of writing $\Sigma^*$. Plus usually stands in for union when dealing with regular expressions, and I'm guessing that's the intention here as well, though it would be well to confirm this.

Assume $L_2$ were context free. By the pumping lemma for context-free languages, we could write any word $w \in L_2$ of length at least $p$ as $uvxyz$ where $|vxy| \leq p$, $|vy| &gt; 0$ such that $uv^kxy^kz \in L$ for all $k \geq 0$. Consider the word $(0^p1^p)(1^p0^p)(0^p1^p)(1^p0^p)\in L_2$ (parentheses added for clarity). We consider cases for the assignment of $vxy$:


$vxy$ consists entirely of $0$s from the first (parenthesized) part of the string. Pumping will change the number of $0$s without changing the number of $1$s in the first part, leading to a string that can't possibly be in the language.
$vxy$ consists of a mix of $0$s and $1$s from the first (parenthesized) part of the string. Pumping will change the numbers of $0$s and $1$s in the first part without affecting the other parts, leading to a string that cannot be in the language.
$vxy$ consists of a $1$s from the first (parenthesized) part only. See argument in point 1 above.
$vxy$ consists of $1$s from the first part and $0$s from the second. See arguments above.


etc. There are 15 cases in total, but the arguments are all basically the same: you always get some parts out of sync (the $p$s won't match). This means that the language is not context-free.