If $G$ is amenable, then every weakly mixing representation $\pi$ has almost invariant finite-dimensional subspaces. This means that $\pi \otimes \bar \pi$ has almost invariant vectors.

Results like this can be found in

M.E.B. Bekka. *Amenable unitary representations of locally compact groups*. Invent. Math. **100** (1990), 383–401.

A concrete example of a weakly mixing representation without almost invariant vectors would be given by the Heisenberg group $H(\mathbb Z)$ acting on $\ell^2(\mathbb Z^2)$ with generator of the center acting by multiplication with some irrational $\theta \in S^1$. It is easy to see that there are no finite-dimensional subrepresentations and since the generator of the center acts by $\theta$ (and not by $1$), there cannot be and almost invariant vectors.