About the turing machine's non-deterministic branches
I guess, with branches you mean the computation branches (the sequence of the konfigurations M has in its computation) of $M$. Then in general a NTM $M$ will accept an input x, if there exists one computation branch in the computation tree, which leads to accept x, no matter, where other branches may lead to (even if there infinite long). So, Machine 1 and 3 will accept. 

I am definitely not sure about the following: I guess, the question, what Machine 2 will do is not decidable (please correct me, if I am wrong)