Solution:

Every function is $1$-fine, so the answer to both parts of the problem is affirmative. Let us further assume $k \geqslant 2$. Every $k$-fine function is $1$-$1$ because from $f(m)=f(n)$ it follows that $m^{k}=f_{k}(m)=f_{k}(n)=n^{k}$, i.e., $m=n$.

a) Answer: YES. Let us construct the function $f$ inductively in the following way. Let $n$ be the smallest natural number whose image has not been determined.

(1) if $n=1$, then $f(n)=1$;

(2) if $n=a^{k}$ for some integer $a>1$, we define $f(n)=f(a)^{k}$;

(3) if $n$ is not a perfect $k$-th power, we choose the smallest $k-1$ natural numbers $n_{1}, n_{2}, \ldots, n_{k-1}$ which are not perfect $k$-th powers and whose images have not yet been determined, and we define $f\left(n_{1}\right)=n_{2}, f\left(n_{2}\right)=n_{3}, \ldots, f\left(n_{k-1}\right)=n_{1}^{k}$.

In this way the function $f$ is well defined. Let us show that it is $k$-fine. For every $n \in \mathbb{N}$ which is not a $k$-th power there exist numbers $n_{1}, \ldots, n_{k-1}$ from condition (3) such that $n_{i}=n$ for some $1 \leqslant i \leqslant k-1$. Then it holds that $f_{k}\left(n_{i}\right)=f_{i}\left(n_{1}^{k}\right)=f_{i}\left(n_{1}\right)^{k}=f\left(n_{i}\right)^{k}$. Also, if $n$ is a perfect $k$-th power, then $n=n_{i}^{k^{s}}$ for some $i$ and $s$, so according to (2) it holds that $f_{k}(n)=f_{k}\left(n_{i}\right)^{k^{s}}=n_{i}^{k^{s+1}}=n^{k}$, which proves our claim.

b) Answer: NO. Indeed, if $f$ is onto and $k$-fine, then for every $a_{0}$ there exists a sequence of natural numbers $a_{1}, a_{2}, \ldots$ such that $f\left(a_{k+1}\right)=a_{k}$ for all $k$, from which $a_{k}^{k}=f_{k}\left(a_{k}\right)=a_{0}$, which is impossible if $a_{0}$ is not a $k$-th power.