Solution:
$x$ must not satisfy $0 \leq x^{2}-x-2<1 \Rightarrow x$ must not satisfy $x^{2}-x-2=(x-2)(x+1) \geq 0 \Leftrightarrow x \in(-\infty,-1] \cup[2,+\infty)$ AND must not satisfy $x^{2}-x-2 \leq 1 \Leftrightarrow x^{2}-x-3<0 \Leftrightarrow x \in\left(\frac{1-\sqrt{13}}{2}, \frac{1+\sqrt{13}}{2}\right)$. Thus, $x \in\left(-\infty, \frac{1-\sqrt{13}}{2}\right] \cup(-1,2) \cup\left[\frac{1+\sqrt{13}}{2},+\infty\right)$.