a) Yes, f(n) = 2n should do fine.
b) Think about doing something similar to (a), but in reverse: If n is odd, so n = 2m + 1, let f(n) = m; if n is even, so n = 2m, let f(n) = m. Because the set of naturals is infinite, you can always get to a given value in N, eventually, by taking a large-enough value 2m or 2m + 1. But obviously having two values (namely, 2m and 2m + 1) go to one value (namely, m) makes this not one-to-one.
(I'm sure there are many, many other ways to answer this part. The above is just what happened to occur to me first.)
c) A constant function is a great choice for this!