Three, (50 points) Solve the system of equations:
$$
\left\{\begin{array}{l}
x-y+z-w=2 \\
x^{2}-y^{2}+z^{2}-w^{2}=6 \\
x^{3}-y^{3}+z^{3}-w^{3}=20 \\
x^{4}-y^{4}+z^{4}-w^{4}=66
\end{array}\right.
$$
Let $p=x+z, q=x z$, then we have
$$
\begin{array}{l}
p^{2}=x^{2}+z^{2}+2 q, \\
p^{3}=x^{3}+z^{3}+3 p q, \\
p^{4}=x^{4}+z^{4}+4 p^{2} q-2 q^{2} .
\end{array}
$$

Similarly, let $s=y+w, t=y w$, then we have
$$
\begin{array}{l}
s^{2}=y^{2}+w^{2}+2 t, \\
s^{3}=y^{3}+w^{3}+3 s t \\
s^{4}=y^{4}+w^{4}+4 s^{2} t-2 t^{2} .
\end{array}
$$

Under this notation system, the first equation of the original system is
$$
p=s+2 \text{. }
$$

Thus, $p^{2}=s^{2}+4 s+4$,
$$
\begin{array}{l}
p^{3}=s^{3}+6 s^{2}+12 s+8, \\
p^{4}=s^{4}+8 s^{3}+24 s^{2}+32 s+16 .
\end{array}
$$

Substituting the expressions for $p^{2}, p^{3}, p^{4}$ and $s^{2}, s^{3}, s^{4}$, we get
$$
\begin{array}{l}
x^{2}+z^{2}+2 q=y^{2}+w^{2}+2 t+4 s+4, \\
x^{3}+z^{3}+3 p q=y^{3}+w^{3}+3 s t+6 s^{2}+12 s+8 \\
x^{4}+z^{4}+4 p^{2} q-2 q^{2} \\
=y^{4}+w^{4}+4 s^{2} t-2 t^{2}+8 s^{3}+24 s^{2}+32 s+16 .
\end{array}
$$

Using the last three equations of the original system to simplify, we get
$$
\begin{array}{l}
q=t+2 s-1, \\
p q=s t+2 s^{2}+4 s-4, \\
2 p^{2} q-q^{2}=2 s^{2} t-t^{2}+4 s^{3}+12 s^{2}+16 s-25 .
\end{array}
$$

Substituting equations (1) and (2) into equation (3), we get
$$
t=\frac{s}{2}-1 \text{. }
$$

Substituting equation (5) into equation (2), we get
$$
q=\frac{5}{2} s-2 \text{. }
$$

Substituting equations (1), (5), and (6) into equation (4), we get $s=2$.
Thus, $t=0, p=4, q=3$.
Therefore, $x, z$ and $y, w$ are the roots of the equations $x^{2}-4 x+3=0$ and $y^{2}-2 y=0$, respectively, i.e.,
$$
\left\{\begin{array}{l}
x=3, \\
z=1
\end{array} \text{ or } \left\{\begin{array}{l}
x=1, \\
z=3,
\end{array} \text{ and } \left\{\begin{array}{l}
y=2, \\
w=0
\end{array} \text{ or } \left\{\begin{array}{l}
y=0, \\
w=2.
\end{array}\right.\right.\right.\right.
$$

In summary, the system of equations has the following four solutions:
$$
\begin{aligned}
(x, y, z, w)= & (3,2,1,0),(3,0,1,2), \\
& (1,2,3,0),(1,0,3,2) .
\end{aligned}
$$

Note: If only one solution or an incomplete set is obtained, the maximum score is 40 (provided by Jia Yuyu).