If $L1⊆L2$, and $L1\not∈RE$, is it possible that $L2∈RE$
Yes. Actually, it is even true that for every language $L_1$ there is a language $L_2 \supseteq L_1$ such that $L_2$ is regular.

Proof: Pick $L_2 = \Sigma^*. \square$