If $A \cap B$ or $A \cup B$ or $A \times B$ is recursively enumerable is it true to say that both $A$ and $B$ are recursively enumerable?
Notice that $\mathsf{RE}$ is not closed under complementation. Therefore there exists an $A\in\mathsf{RE}$ such that $B=A^c\notin\mathsf{RE}$, while $A\cap B$ and $A\cup B$ are trivially recursively enumerable.

However, when $A\times B\in\mathsf{RE}$, there is a Turing machine $\mathcal{M}$ accepts $(x,y)$ iff $x\in A$ and $y\in B$. Therefore $\mathcal{M}(x,y_0)$ for fixed $y_0\in B$ accepts the language $A$, and hence $A\in\mathsf{RE}$. For $B$ it is the same.

Update: I notice that I missed something in the case of $A\times B$. The conclusion holds only when both $A,B$ are not empty, since otherwise $A\times B=\varnothing\in\mathsf{RE}$ whatever the other one is.