1). REG = {<M> | L(M) is a regular set }
Rice second theorem can help here =>
Take Tyes = {0,1,10,01} and TNo = (0,1)*
Now, as Tyes is a proper subset of TNo , hence, making this language even not RE.
2). Complement of REG
Apply rice second theorem again =>
Take Tyes = Any non regular language like CFL or CSL and TNo = (0,1)*
Now, again as Tyes is a proper subset of TNo , hence, making this language even not RE.