For the second, here is a counterexample: Consider

Now, and is thus finite, which is easy to show by the squeeze/sandwich theorem, using and

But, is nowhere continuous. To show this, choose any

First, if is rational, then . Take Now, no matter how small you choose to be, there is always going to be an irrational number within units of in which case Thus, and is not continuous at The case for irrational should be similar.

And of course, if is everywhere discontinuous, then at any given value its derivative is undefined.

Does that help any? I could probably get you started on (a), but you'll have to wait a bit.