1. Set of strings common in $L$ and $R' = L \cap R'$.
Now recursively enumerable languages are not closed under complement- meaning the complement of a r.e. language may or may not be r.e. So, lets say it is not r.e. Now, let our regular language $L$ be $\Sigma^*$. So, we get $\Sigma^* \cap A = A$, where $A = R'$ is a not r.e. language.
2. Possible. One such case is if we take $L = \emptyset$.
So, C is the answer here.