Minimum number of vertices whose removal makes the graph an independent set
Yes, and the algorithm is straight forward branching.

Suppose that you are given a graph $G$ and an integer $k$.  If $k = 0$, we simply return whether $G$ is an independent set.  If $G$ is an independent set, we return "yes".  Otherwise, we branch.  If there is an edge $uv$ in $G$, we recurse on $G-u, k-1$ and $G-v, k-1$.  Since the budget drops with one for every recursive call, we call ourselves at most $2^k$ times.

The correctness is left as an exercise.