Mixing cobalt(III) hydroxide and thallium(III) hydroxide
This should be a comment, but I decided to post it as an answer, since from my point of view, the present answers fail to shed light on the absolute simplicity of the truth.
We can state from the present answers that there will be no precipitates, but not because the concentration is smaller than an atom per liter, we don't know the volume used since it was not stated.
The simple fact is that the $[\ce{OH^-}]$ concentration stays the same, it will be $10^{-7}$ before and after the mixing of the solutions, no matter how much from each solution we will mix together, since the solubility product is so small and the dominating factor for the $[\ce{OH^-}]$ concentration comes from the autoprotolysis of water. Accepting this little knowledge, no matter what we will never reach the saturation concentration after mixing. The saturation concentration will stay the same always (in this problem), so we will just dilute the initial concentration.
Imagine we mix an equal amount of both solutions, then the before saturated concentration of the substances will be halved. So in the end in this specific case for an example the solution will be half saturated with both hydroxides. That is why there is no precipitation.
Edit: Also there is no sense talking about common ion effect, since this phenomenon arises from the charge balance equation.
$$3[\ce{Tl^{3+}}]+3[\ce{Co^{3+}}]+[\ce{H^+}]=[\ce{OH^-}]$$
As we postulated before $[\ce{H^+}]=[\ce{OH^-}]$, since the autoprotolysis is the domination factor, both of the concentration of the ions will be zero (simplified). In truth the $[\ce{OH^-}]$ concentration will be just slightly higher than the $[\ce{H^+}]$ concentration, but we can neglect this equation since we assumed the only $[\ce{OH^-}]$ concentration comes from autoprotolysis, meaning we don't have to solve a system of equations, just plugging in the value in the solubility product.