If $L1⊆L2$, and $L1\not∈RE$, is it possible that $L2∈RE$ If $L1⊆L2$, and $L1\not∈RE$, is it possible that $L2∈RE$ ?

Also I find it hard to find languages that are not in RE at all, I've heard about Arithmetical hierarchy but we didn't really learnt it in class so I guess the solution shouldn't necessarily be correlated to Arithmetical hierarchy.

I know that the $A_{TM}=\{(〈M〉,w): M\ accepts\ w\}$ where M is a Turing machine
then $\overline {A_{TM}}\notin RE$, but it doesn't seem to help.
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 $L_2 = \Sigma^*. \square$