Let $f_1,...,f_n$ be an enumeration of all computable functions in N to N. Prove $h(n)=f_n(n)$ is not calculable. Let $f_1,...,f_n$ be an enumeration of all calculable functions in N to N. Prove $h(n)=f_n(n)$ is not calculable.
I don't really know where to start here, any hint appreciated.

Hint : Let us assume $h$ is calculable. Then, so is $1+h$. This means $\exists n_0 \in \mathbb{N}, f_{n_0}=1+h$. What is $h(n_0)$?
