Solution:
The inequality is defined for $x^{2}-x-2>0$ and $3+2x-x^{2}>0$, whence $x \in (2,3)$.

Since $x=a+1$ is a solution, we have $a \in (1,2)$.

Then the inequality is equivalent to
$$x^{2}-x-2 > 3+2x-x^{2} \Longleftrightarrow (x+1)(2x-5)>0$$
and therefore $x \in \left(\frac{5}{2}, 3\right)$.