Can the intersection two non-recursive sets be recursive? Prove it
Short answer: there are such languages.
Take some language $L\notin RE\cup co-RE$ (look at the explanation at the end for why such a language exists). Then both $L$ and $\bar L$ are not recursive, and are not even recursively enumerable. But, its obvious that $\emptyset=L\cap \bar L$ is a recursive set.
Claim: there is some $L\notin RE\cup co-RE$.
Proof:
Why would such $L$ exist? We will compare the sizes of $RE\cup co-RE$ and the set of all languages.
$RE$ contains only $\aleph_0$ languages, since each language has to be recognized by some turing machine, and turing machines we can ultimately be encoded in text. More formally, we know that for any $A\in RE$ we must have some turing machine $T$ that recognizes it, and $\langle T \rangle\in\{0,1\}^{\mathbb{N}}$. But $|\{0,1\}^\mathbb{N}|=\aleph_0$, and thus $|RE|\le\#\text{TM}\le|\{0,1\}^{\mathbb{N}}|=\aleph_0$. A similar argument can be done to show that also $|co-RE|=\aleph_0$ and combining the two, we get $|RE\cup co-RE| = \aleph_0$
Now, notice that the set of all languages, is precisely $P(\{0,1\}^{\mathbb{N}})$, and thus it has $2^{\aleph_0}$ elements in it. Since $2^{\aleph_0}&gt;\aleph_0$, there must be some $L\in P(\{0,1\}^{\mathbb{N}})$ such that $L\notin RE\cup co-RE$, and thus we proved the claim.