a) Just take $f(x)=a \cdot x^b$ where $b$ is the positive solution of $b^2=b+1$ and $a$ is appropriate. b) Clearly $f'$ is injective so that $f$ takes each value at most twice. Since $f$ is surjective, it must in fact be bijective (otherwise takes some value at least three times) and so $f'$ must also be bijective. But on the other hand, if $f$ is bijective, it is monotone and hence $f'$ cannot be surjective. Contradiction!