Three. (16 points) As shown in Figure 3, the side length of square $ABCD$ is 1, and points $M$ and $N$ are on $BC$ and $CD$ respectively, such that the perimeter of $\triangle CMN$ is 2. Find:
(1) The size of $\angle MAN$;
(2) The minimum area of $\triangle MAN$.
(1) As shown in Figure 4, extend $C B$ to $L$ such that $B L = D N$, then
Rt $\triangle A B L \cong$ Rt $\triangle A D N$.
Thus, $A L = A N, \angle 1 = \angle 2$,
$$
\begin{array}{l}
\angle N A L = \angle D A B = 90^{\circ}. \\
\text{Also, } M N = 2 - C N - C M \\
= D N + B M \\
= B L + B M = M L,
\end{array}
$$

and $A M = A M$, so
$\triangle A M N \cong \triangle A M L$. Therefore, $\angle M A N = \angle M A L = \frac{90^{\circ}}{2} = 45^{\circ}$.
(2) Let $C M = x, C N = y, M N = z$. Then
$$
\left\{\begin{array}{l}
x + y + z = 2, \\
x^2 + y^2 = z^2
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x = 2 - y - z, \\
x^2 + y^2 = z^2 .
\end{array}\right.\right.
$$

Thus, $(2 - y - z)^2 + y^2 = z^2$.
Simplifying, we get $2 y^2 + (2 z - 4) y + (4 - 4 z) = 0$.
Since $y > 0$, we have
$$
\Delta = 4(z - 2)^2 - 32(1 - z) \geqslant 0 \text{, }
$$

which simplifies to $(z + 2 + 2 \sqrt{2})(z + 2 - 2 \sqrt{2}) \geqslant 0$.
Since $z > 0$, it follows that $z \geqslant 2 \sqrt{2} - 2$.
Equality holds if and only if $x = y = 2 - \sqrt{2}$.
Since $S_{\triangle M V V} = S_{\triangle M L}$
$$
= \frac{1}{2} \cdot M L \cdot A B = \frac{1}{2} M N \times 1 = \frac{z}{2} \text{, }
$$

therefore, the minimum area of $\triangle A M N$ is $\sqrt{2} - 1$.