13. 3.17 ** Let the equation of the circumcircle of square $ABCD$ be $x^{2}+y^{2}-6x+a=0$ ($a<9$), and the slope of the line $l$ on which points $C$ and $D$ lie is $\frac{1}{3}$.
(1) Find the coordinates of the center $M$ of the circumcircle and the slopes of the diagonals $AC$ and $BD$ of the square;
(2) If points $A$ and $B$ in the upper half of the $x$-axis lie on a parabola with the origin as its vertex and the $x$-axis as its axis of symmetry, find the equation of this parabola and the equation of line $l$.
(1) As shown in the figure, given $(x-3)^{2}+y^{2}=9-a(a>0)$. Since points $A$ and $B$ lie on the parabola, then
$$
\left\{\begin{array}{l}
\left(\frac{\sqrt{5}}{5} r\right)^{2}=2 p\left(3-\frac{2 \sqrt{5}}{5} r\right), \\
\left(\frac{2 \sqrt{5}}{5} r\right)^{2}=2 p\left(3+\frac{\sqrt{5}}{5} r\right)
\end{array}\right.
$$

Solving these equations, we get $r=\sqrt{5}, p=\frac{1}{2}$, so the equation of the parabola is $y^{2}=x$. From this, we know that the coordinates of point $A$ are $(1,1)$, and the coordinates of point $C$, which is the reflection of point $A$ about point $M(3,0)$, are $(5,-1)$. The equation of line $l$ is $y-(-1)=\frac{1}{3}(x-5)$, which simplifies to $x-3 y-8=0$.