Yes $ L1⊆L2 $and $ L1 not∈RE $, It is possible that $ L2∈RE $ ?

It's also hard for me to find languages that aren't RE at all, I've heard of arithmetic hierarchy, but we didn't really learn it in class, so I guess the solution doesn't necessarily have to be correlated with arithmetic hierarchy.

I know that he $ A_ {TM} = {(〈M〉, w): M accepts w } $ where M is a Turing machine

so $ overline {A_ {TM}} notin RE $, but it doesn't seem to help.