Complexity of Two problems graphs with cycle
The first problem can be answered by finding the length of the shortest cycle in $G$, a parameter known as the girth of $G$. Indeed, if $G$ has girth $g$ then $G$ contains a cycle of length at most $k$ iff $g \leq k$. You can use BFS to find the girth efficiently, see for example these course notes. Therefore this problem can be solved in polynomial time.

The second problem is NP-complete by reduction from Hamiltonian cycle, as you mention (just choose $k = n$).