13. Given the vectors $\boldsymbol{a}=(1, x), \boldsymbol{b}=\left(x^{2}+x,-x\right)$, and a constant $m$ such that $m \leqslant-2$, find the range of $x$ that satisfies $\boldsymbol{a} \cdot \boldsymbol{b}+2>$ $m\left(\frac{2}{\boldsymbol{a} \cdot \boldsymbol{b}}+1\right)$.
13. Since $\boldsymbol{a}=(1, x), \boldsymbol{b}=\left(x^{2}+x,-x\right)$, we have $\boldsymbol{a} \cdot \boldsymbol{b}=x^{2}+x-x^{2}=x$; thus, the inequality $\boldsymbol{a} \cdot \boldsymbol{b}+2>$ $m\left(\frac{2}{a \cdot b}+1\right)$ is equivalent to $x+2>m\left(\frac{2}{x}+1\right)$. Simplifying this, we get $\frac{(x+2)(x-m)}{x}>0$,
which is $x(x+2)(x-m)>0$. We can now discuss the cases:
(1) If $m=-2$, the range of $x$ is $(0,+\infty)$;
(2) If $m<-2$, the range of $x$ is $(m,-2) \cup(0,+\infty)$.