Three. (20 points) Let the function $f(x)=a^{x}+3 a(a>$ 0 , and $a \neq 1)$ have the inverse function $y=f^{-1}(x)$, and the graph of the function $y=g(x)$ is symmetric to the graph of $y=f^{-1}(x)$ about the point $(a, 0)$.
(1) Find the analytical expression of the function $y=g(x)$.
(2) Does there exist a real number $a$ such that when $x \in[a+2, a+3]$, it always holds that $\left|f^{-1}(x)-g(-x)\right| \leqslant 1$? If it exists, find the range of $a$; if not, explain the reason.
(1) From $f(x)=a^{x}+3 a$, we get $f^{-1}(x)=\log _{a}(x-3 a)(x>3 a)$.
Also, the graph of $y=g(x)$ is symmetric to the graph of $y=f^{-1}(x)$ about the point $(a, 0)$, thus we have
$$
g(x)=-\log _{a}(-x-a)(x3 a),
$$

which means $-1 \leqslant \log _{a}\left(x^{2}-4 a x+3 a^{2}\right) \leqslant 1$.
From $3 a0$, we get $0<a<1$.
Therefore, $a \leqslant x^{2}-4 a x+3 a^{2} \leqslant \frac{1}{a}$, which is
$$
\left\{\begin{array}{l}
x^{2}-4 a x+3 a^{2}-a \geqslant 0, \\
x^{2}-4 a x+3 a^{2}-\frac{1}{a} \leqslant 0 .
\end{array}\right.
$$

Solving equation (1) yields
$$
x \leqslant 2 a-\sqrt{a^{2}+a} \text { or } x \geqslant 2 a+\sqrt{a^{2}+a} \text {. }
$$

By the problem's condition,
$$
\begin{array}{r}
{[a+2, a+3] \subseteq\left(-\infty, 2 a-\sqrt{a^{2}+a}\right] \cup} \\
{\left[2 a+\sqrt{a^{2}+a},+\infty\right) .}
\end{array}
$$

Thus, $a+3 \leqslant 2 a-\sqrt{a^{2}+a}$ or
$$
a+2 \geqslant 2 a+\sqrt{a^{2}+a} \text {. }
$$

Combining $0<a<1$, we solve to get $0<a \leqslant \frac{4}{5}$.
For equation (2), let $h(x)=x^{2}-4 a x+3 a^{2}-\frac{1}{a}$.
Then $[a+2, a+3]$ is a subset of the solution set of the inequality $h(x) \leqslant 0$ if and only if
$$
\left\{\begin{array}{l}
h(a+2)=-\frac{1}{a}(2 a-1)^{2} \leqslant 0, \\
h(a+3)=-\frac{1}{a}\left(6 a^{2}-9 a+1\right) \leqslant 0 .
\end{array}\right.
$$

Combining $0<a<1$, we solve to get $0<a \leqslant \frac{9-\sqrt{57}}{12}$.
In summary, there exists $0<a \leqslant \frac{9-\sqrt{57}}{12}$, such that when $x \in$ $[a+2, a+3]$, the inequality $\left|f^{-1}(x)-g(-x)\right| \leqslant 1$ always holds.