Complement of HAMPATH
The HAMPATH complement ("G does not contain an Hamiltonian path from s to t") is in co-NP; to be more precise it is  co-NP complete (it is easy to prove that $L$ is NP-complete if and only if its complement $\bar{L}$ is co-NP-complete).

The question if it is in NP is open.

But since HAMPATH is NP-complete, if its complement is in $\mathsf{NP}$ then $\mathsf{NP} = \mathsf{co\mbox{-}NP}$