The trivial category $1$ has only one object $*$, and the only arrow is the identity arrow. Is $1$ cocomplete ? In other words, does any functor $\mathbb{C}\overset{F}\to 1$ from a small category $\mathbb{C}$ have a colimit ?
If not I would like to see a functor $F$ with no colimit.
On
Yes, the terminal category is cocomplete. The terminal category is equivalent to the presheaf category $[\emptyset^{op}, \mathcal{Set}]$, where $\emptyset$ is the empty category. As presheaf categories are always cocomplete (and complete), the terminal category is cocomplete.
You could also use the definition of a colimit directly. Since the hom sets are all trivial and there's only one possibility for the colimit, this should be fairly easy.
Yes: the colimit of any diagram is given by the unique object. More abstractly, a category $\mathbb A$ has $\mathbb C$-indexed colimits if and only if the diagonal functor $\Delta : \mathbb A \to \mathbb A^{\mathbb C}$ has a left adjoint. When $\mathbb A = 1$, the diagonal functor is an equivalence, so trivially has a left adjoint.