It is still not clear what is claimed about L(M1) L(M2). I'll assume it says L(M1) is a subset of L(M2), judging from SUBSETTM.

For 1, the easiest thing is to use

Rice's theorem, which says that if you have a nontrivial set of

*languages*, then the set of (codes of) TMs that accept these languages is not decidable. In this problem, however, we have

*pairs* of languages. One trick is to specialize the problem to make it fit Rice's theorem by fixing M2. For example, fix some M2 such that L(M2) is empty. In this case, if SUBSETTM is decidable, then its subset with fixed M2 {<M1, M2> : M1 is a Turing machine and L(M1) is empty} is decidable as well. This, in turn, means that {<M> : M is a Turing machine and L(M1) is empty} is decidable.

One of the equivalent conditions (if not the definition) of recognizable language is that some TM stops on strings in the language and does not stop on strings not in the language. Well, given a (code of a) machine M, one can start running it on all possible strings. One can run it n steps on the m'th string and enumerate pairs (n, m) to try all of them (this is similar to the proof that

is countable). One will eventually learn

*if* the machine accepts two different strings, but may never learn if it does not.