1. Let the real number $a$ satisfy $a<9 a^{3}-11 a<|a|$, then the range of values for $a$ is $\qquad$
Answer: $a \in\left(-\frac{2 \sqrt{3}}{3},-\frac{\sqrt{10}}{3}\right)$.
Solution: From $a\frac{9 a^{3}-11 a}{a}>\frac{|a|}{a}=-1$, 
$$
it follows that $-1<9 a^{2}-11<1$, so $a^{2} \in\left(\frac{10}{9}, \frac{4}{3}\right)$. Since $a<0$, we have $a \in\left(-\frac{2 \sqrt{3}}{3},-\frac{\sqrt{10}}{3}\right)$.