jmc

First, we simplify both of the congruence relationships to get: $$\begin{array}{rl}& x-1\equiv 1-x(\mathrm{mod}12)⟹2x\equiv 2(\mathrm{mod}12),\\ & x-2\equiv 2-x(\mathrm{mod}12)⟹2x\equiv 4(\mathrm{mod}12).\end{array}$$Since $2x$ cannot be equivalent to both 2 and 4 mod 12, we know that there are $\boxed{0}$ solutions.