9. (16 points) Given the function $f(x)=\left|x^{2}-a\right|$, where $a$ is a positive constant. If there are exactly two pairs of solutions $(m, n)$ such that the range of $f(x)$ on the domain $[m, n]$ is also $[m, n]$, find the range of values for $a$.

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II. 9. Since $f(x)=\left|x^{2}-a\right| \geqslant 0$, therefore, $n>m \geqslant 0$.
(1) It is easy to find that the x-coordinate of the intersection point of $y=x$ and $y=x^{2}-a(x \geqslant \sqrt{a})$ is $x=\frac{1+\sqrt{1+4 a}}{2}$.
Obviously, the interval
$[m, n]=\left[0, \frac{1+\sqrt{1+4 a}}{2}\right]$ satisfies the requirement.
(2) Since $f(0)=a$, if $[m, n]=[0, a]$, then, from $x^{2}-a=a \Rightarrow x=\sqrt{2 a}$.
According to the problem, we only need $\sqrt{a} \leqslant a \leqslant \sqrt{2 a} \Rightarrow 1 \leqslant a \leqslant 2$.
However, when $a=2$, $\frac{1+\sqrt{1+4 a}}{2}=a$, there is only one solution, so $1 \leqslant a<2$.
If $a>0$, then $n>0$, \\ g(\sqrt{a})=1-\sqrt{a}>0, \\ \Delta=1-4(1-a)>0 .\end{array}\right.$
Solving this gives $\frac{3}{4}<a<1$.
In summary, the range of values for $a$ is $\frac{3}{4}<a<2$.