1. The side length of square $A B C D$ is 4. A semicircle is drawn inside the square with $A B$ as its diameter. $C M$ and $D N$ are tangents to the semicircle, with $M$ and $N$ being the points of tangency. If $C M$ and $D N$ intersect at a point $P$ inside the square, then the area of $\triangle P M N$ is
$$
\text { II. } 1 . \frac{27}{25} \text {. }
$$

As shown in Figure 3, let $C M$ intersect $A D$ at point $E$, and $D N$ intersect $B C$ at point $F$. Then quadrilateral $E F C D$ is a rectangle.
By the equal tangent segments, we have
$$
C M=C B=4 \text {. }
$$

Let $x=E A=E M$, then
$$
D E=4-x, C E=4+x \text {. }
$$

In the right triangle $\triangle C D E$, by the Pythagorean theorem, we have
$$
(4-x)^{2}+4^{2}=(4+x)^{2} \text {. }
$$

Solving for $x$ gives $x=1$.
Therefore, $D E=3, C E=5$.
Thus, $S_{\text {Rectangle EFCD }}=12$. So,
$$
S_{\triangle P E F}=\frac{S_{\text {Rectangle EFCD }}}{4}=3 \text {. }
$$

By symmetry, $M N / / E F$, then
$$
P M=P E-x=\frac{5}{2}-1=\frac{3}{2} \text {. }
$$

And $\frac{S_{\triangle P M N}}{S_{\triangle P E F}}=\left(\frac{P M}{P E}\right)^{2}=\left(\frac{3}{5}\right)^{2}=\frac{9}{25}$, so,
$$
S_{\triangle P M N}=\frac{9}{25} S_{\triangle P E F}=\frac{27}{25}
$$