(A) is true, since we can prove it. Assume that H(M) is decidable, so there exists a deciding TM for it, M1.

We can build a decider for L(M) - on every input x, run M1 on x. If M halts on x, run M on x and accept iff M accepts. otherwise, reject.

(B) is not true, since the proof gets complicated. Assume that L(M) is decidable. Lets say it is the empty set, which is indeed decidable.

Does M really halts on every input? What if M is the machine that for every input enters an infinite loop? Its language will still be decidable (there exists some TM that decides it, not necessarily M), but it loops on every input.

You can find a less trivial example, but that's the explanation, I think.