Three, (25 points) Given that $m, n, p, q$ satisfy $m n p q = 6(m-1)(n-1)(p-1)(q-1)$.
(1) If $m, n, p, q$ are all positive integers, find the values of $m, n, p, q$;
(2) If $m, n, p, q$ are all greater than 1, find the minimum value of $m+n+p+q$.
(1) Let's assume $m \geqslant n \geqslant p \geqslant q$.
Clearly, $q \geqslant 2$.
If $q \geqslant 3$, then
$$
\begin{aligned}
\frac{1}{m} & \leqslant \frac{1}{n} \leqslant \frac{1}{p} \leqslant \frac{1}{q} \leqslant \frac{1}{3} \\
\Rightarrow & \frac{1}{6}=\left(1-\frac{1}{m}\right)\left(1-\frac{1}{n}\right)\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right) \\
& \geqslant\left(\frac{2}{3}\right)^{4}>\frac{1}{6}
\end{aligned}
$$

Contradiction.
Thus, $q=2$.
Substituting $q=2$ into the given equation, we get
$$
\begin{array}{l}
m n p=3(m-1)(n-1)(p-1) . \\
\text { Clearly, } p \geqslant 2 . \\
\text { If } p \geqslant 4, \text { then } \\
\frac{1}{m} \leqslant \frac{1}{n} \leqslant \frac{1}{p} \leqslant \frac{1}{4} \\
\Rightarrow \frac{1}{3}=\left(1-\frac{1}{m}\right)\left(1-\frac{1}{n}\right)\left(1-\frac{1}{p}\right) \\
\quad \geqslant\left(\frac{3}{4}\right)^{3}>\frac{1}{3},
\end{array}
$$

Clearly, $p \geqslant 2$.
If $p \geqslant 4$, then
$$
\begin{aligned}
\frac{1}{m} & \leqslant \frac{1}{n} \leqslant \frac{1}{p} \leqslant \frac{1}{4} \\
\Rightarrow & \frac{1}{3}=\left(1-\frac{1}{m}\right)\left(1-\frac{1}{n}\right)\left(1-\frac{1}{p}\right) \\
& \geqslant\left(\frac{3}{4}\right)^{3}>\frac{1}{3},
\end{aligned}
$$

Contradiction.
Thus, $p=2$ or 3.
When $p=2$, we have
$$
\begin{aligned}
2 m n & =3(m-1)(n-1) \\
\Rightarrow & (m-3)(n-3)=6 \times 1=3 \times 2 \\
& \Rightarrow(m, n)=(9,4),(6,5) .
\end{aligned}
$$

When $p=3$, we have
$$
\begin{array}{l}
m n=2(m-1)(n-1) \\
\Rightarrow(m-2)(n-2)=2 \times 1 \\
\Rightarrow(m, n)=(4,3) . \\
\text { Therefore, }(m, n, p, q) \\
=(9,4,2,2),(6,5,2,2),(4,3,3,2) .
\end{array}
$$
(2) Note that,
$$
\begin{array}{l}
\frac{1}{6}=\left(1-\frac{1}{m}\right)\left(1-\frac{1}{n}\right)\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right) \\
\leqslant\left[\frac{\left(1-\frac{1}{m}\right)+\left(1-\frac{1}{n}\right)+\left(1-\frac{1}{p}\right)+\left(1-\frac{1}{q}\right)}{4}\right] .
\end{array}
$$

That is,
$$
\begin{array}{l}
\frac{1}{m}+\frac{1}{n}+\frac{1}{p}+\frac{1}{q} \leqslant 4-\frac{4}{\sqrt[4]{6}} . \\
\text { Also, }(m+n+p+q)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}+\frac{1}{q}\right) \geqslant 16 \\
\Rightarrow m+n+p+q \\
\quad \geqslant \frac{16}{\frac{1}{m}+\frac{1}{n}+\frac{1}{p}+\frac{1}{q}} \geqslant \frac{16}{4-\frac{4}{4}}=\frac{4 \sqrt[4]{6}}{\sqrt[4]{6}-1} .
\end{array}
$$

Equality holds if and only if $m=n=p=q=\frac{\sqrt[4]{6}}{\sqrt[4]{6}-1}$.
Thus, $(m+n+p+q)_{\min }=\frac{4 \sqrt[4]{6}}{\sqrt[4]{6}-1}$.