Intersection/Union of recursively enumerable languages that aren't decidable?
Correct. We can have $L_1\cap L_2=\emptyset\in R$. A good answer would give an example of such a pair $L_1$, $L_2$ so you should figure that out on your own.
Correct but for the wrong reason. If $L_1$ and its complement are both $RE$, then both are recursive and the question says that neither is recursive.  You should prove this yourself, as you'll need it for the next part.


Hint for this proof: show how to use machines that accept $L_1$ and $\overline{L_1}$ to give a machine that decides&nbsp;$L_1$.)
Hint for the question: take a recursives language $K_1$ and $K_2$ and set $L_1=K_1 \cup \text{something}$ and $L_2=K_2\cup\text{something else}$ in a way that gives $L_1\cup L_2=K_1\cup K_2$.

Correct. We know that $L_1\in RE$. Use $L_1\cap L_2\in R$ and $L_1\cup L_2\in R$ to show that $\overline{L_1}\in RE$. By the extra proof I suggested you do in the previous part, that means that $L_1\in R$, contradicting the requirement that $L_1\in RE\setminus R$.