1. Given the equation
$$
\left|x^{2}-1\right|-a x^{2}=2-(a+2) x
$$

has exactly three solutions. Then the range of values for $a$ is $\qquad$
1.0 < a < 1 or $1 < a < 4$.
From the given equation, we have
$$
\begin{array}{l}
\left|x^{2}-1\right|=a x^{2}-(a+2) x+2 \\
=(a x-2)(x-1) .
\end{array}
$$

Obviously, $x=1$ is a root of equation (1).
When $x \neq \pm 1$, equation (1) can be transformed into
$$
\frac{\left|x^{2}-1\right|}{x-1}=a x-2 \text{. }
$$

By combining geometric and algebraic analysis, we know $0 < a < 1$ or $1 < a < 4$.