Three. (20 points) Let the function $f(x)$ satisfy
$$
f\left(2^{x}\right)=x^{2}-2 a x+a^{2}-1 \text {, }
$$

and $f(x)$ has a range of $[-1,0]$ on $\left[2^{a-1}, 2^{a^{2}-2 a+2}\right]$. Find the range of values for $a$.
Let $g(x)=f\left(2^{x}\right)$, then
$f(x)=g\left(\log _{2} x\right)=\left(\log _{2} x\right)^{2}-2 a \log _{2} x+a^{2}-1$.
Thus, the range of $f(x)$ on the interval $\left[2^{a-1}, 2^{a^{2}-2 a+2}\right]$ being $[-1,0]$ is equivalent to the range of $g(x)=x^{2}-2 a x+a^{2}-1$ on the interval $\left[a-1, a^{2}-2 a+2\right]$ being $[-1,0]$.

Since $g(a)=-1$ is the minimum value of $g(x)$ on $\mathbf{R}$, it follows from the condition that $a \in\left[a-1, a^{2}-2 a+2\right]$, and the maximum value of $g(x)$ on this interval must be attained at one of the endpoints; also, the function value of $g(x)$ at the left endpoint $g(a-1)=0$ is precisely the maximum value of $g(x)$ on this interval, so $a$ must be in the right half of the interval, i.e.,
$$
\frac{(a-1)+\left(a^{2}-2 a+2\right)}{2} \leqslant a \leqslant a^{2}-2 a+2 \text {. }
$$

Solving this yields $\frac{3-\sqrt{5}}{2} \leqslant a \leqslant 1$ or $2 \leqslant a \leqslant \frac{3+\sqrt{5}}{2}$.
Note: Since $g(a+1)=g(a-1)=0$, it must be that $a \leqslant$ $a^{2}-2 a+2 \leqslant a+1$. This also leads to the range of $a$.