(13) As shown in the figure, in the Cartesian coordinate system, the circle $M$ with the equation $x^{2}+y^{2}+D x+E y+F=0$ has an inscribed quadrilateral $A B C D$ whose diagonals $A C$ and $B D$ are perpendicular to each other, and $A C$ and $B D$ lie on the $x$-axis and $y$-axis, respectively.
(1) Prove that $F<0$;
(2) If the area of quadrilateral $A B C D$ is 8, the length of diagonal $A C$ is 2, and $\overrightarrow{A B} \cdot \overrightarrow{A D}=0$, find the value of $D^{2}+E^{2}-4 F$.
13 (1) Method one: From the problem, the origin $O$ must be inside the circle $M$, which means the value of the left side of the equation $x^{2}+y^{2}+D x+E y+F=0$ when substituting the point $(0,0)$ is less than 0. Therefore, we have $F<0$.

Method two: From the problem, it is not difficult to find that points $A$ and $C$ are on the negative and positive halves of the $x$-axis, respectively. Let the coordinates of these two points be $A(a, 0)$ and $C(c, 0)$, then we have $a c<0$.

For the circle equation $x^{2}+y^{2}+D x+E y+F=0$, when $y=0$, we get $x^{2}+D x+F=0$. The roots of this equation are the $x$-coordinates of points $A$ and $C$, so $x_{A} x_{C}=a c=F$. Since $a c<0$, it follows that $F<0$.
(2) For the quadrilateral $ABCD$ with perpendicular diagonals, the area $S=\frac{|A C| \cdot|B D|}{2}$. Given $S=8$ and $|A C|=2$, we can find $|B D|=8$.

Since $\overrightarrow{A B} \cdot \overrightarrow{A D}=0$, $\angle A$ is a right angle, so $|B D|=2 r=8 \Rightarrow r=4$. For the circle represented by the equation $x^{2}+y^{2}+D x+E y+F=0$, we know $\frac{D^{2}}{4}+\frac{E^{2}}{4}-F=r^{2}$, so
$$
D^{2}+E^{2}-4 F=4 r^{2}=64 .
$$