turing machine

Question

Answer 1

$L1=non R.E.$=not even partially decidable

$L2=R.E.$

$L1\cap L2$= non R.E.or outside R.E.

Options are not printed correctly

Now, $L1'$ is $\Sigma ^*-non R.E.$=R.E.

i.e.  the answer

Comments

Its right, but I don't understand the options here :P

What they actually need.

---

Can we say, L =L1 $\cap$ L2= {M | M accepts exactly 2016 strings } ??

---

Yes, I think so.