We know that if a TM accepts $\epsilon$ or not is semi-decidable but not decidable (undecidable) (Rice's theorem). So, the given problem must also be undecidable as by taking $M_2 = M_1$, it reduces to the known undecidable problem.
Now, the given problem is semi-decidable, because we can semi-decide it in two steps. First see if $M_1$ accepts $\epsilon$ and then see if $M_2$ accepts $\epsilon.$ If $\epsilon$ is in $L$ both $M_1$ and $M_2$ will halt making our problem semi-decidable.