10. (20 points) As shown in Figure 1, rectangle $O A B C$ is a billiard table, with $O A=12, O C=6$. A ball is hit from point $O$ and can bounce off the four edges of the table infinitely. It is known that the ball passes through the center of rectangle $O A B C$, $D(6,3)$.

Figure 1
(1) Find the number of all integer points $E(a, b)(1 \leqslant a \leqslant 11$, $1 \leqslant b \leqslant 5)$ such that the ball can pass through point $E$;
(2) If the shortest path length of the ball between points $D$ and $E$ is $f(E)$, find the maximum value of $f(E)$.
10. Translate the rectangle $O A B C$ and points $D, E$ as a whole to the right and upward, to get a series of rectangles.

Let the images of $D, E$ in the $i(i=0,1, \cdots)$-th column and $j(j=0,1, \cdots)$-th row of rectangles be $D_{i j}, E_{i j}$. Then
$$
\begin{array}{l}
D_{i j}=(12 i+6,6 j+3) ; \\
E_{2 t, 2 t}=(24 s+a, 12 t+b), \\
E_{2,2 t+1}=(24 s+a, 12 t+12-b), \\
E_{2 t+1 ; 2 t}=(24 s+24-a, 12 t+b), \\
E_{2+1,2 t+1}=(24 s+24-a, 12 t+12-b) .
\end{array}
$$

Let the slopes of $O D_{i j}, O E_{i j}$ be $k_{i j}, l_{i j}$. Then
$$
k_{i j}=\frac{6 j+3}{12 i+6}=\frac{2 j+1}{2(2 i+1)} \text {, }
$$
$l_{2,, 2 t}=\frac{12 t+b}{24 s+a}, l_{2 s, 2 t+1}=\frac{12(t+1)-b}{24 s+a}$,
$$
l_{2 s+1,2 t}=\frac{12 t+b}{24(s+1)-a}, l_{2 s+1,2+1}=\frac{12(t+1)-b}{24 s(s+1)-a} \text {. }
$$

Therefore, the ball passing through point $D$ can also pass through point $E$
$\Leftrightarrow$ there exist $i_{1}, i_{2}, j_{1}, j_{2}$ such that $k_{i v_{1}}=l_{i j j_{2}}$.
For equation (2) to hold, it is necessary that
$$
2^{\alpha+1}\left\|a, 2^{\alpha}\right\| b(\alpha \in\{0,1,2\}) \text {. }
$$

Thus, $E(a, b)=(2,1),(4,2),(6,3),(8,4),(10,5)$; $(2,3),(10,3),(6,1),(6,5)$; $(2,5),(10,1)$.

Below, we use equation (2) to verify that the ball can pass through the above points $E(a, b)$, and calculate $f(E)=f(a, b)$.
For the first five points $E$, we have $k_{00}=l_{00}=\frac{1}{2}$, and
$$
\begin{array}{l}
f(2,1)=f(10,5)=2 \sqrt{5}, \\
f(4,2)=f(8,4)=\sqrt{5}, f(6,3)=0 .
\end{array}
$$

For the middle four points $E$, we have $k_{i j} \neq l_{i j}$, so the ball must bounce off one of the four sides of the billiard table between $D$ and $E$, and
$$
\begin{array}{l}
k_{01}=\frac{3}{2}=l_{00}(2,3)=l_{02}(10,3), \\
k_{10}=\frac{1}{6}=l_{00}(6,1)=l_{20}(6,5) .
\end{array}
$$

Thus, $f(2,3)=f(10,3)=2 \sqrt{13}$,
$$
f(6,1)=f(6,5)=2 \sqrt{37} \text {. }
$$

For $E=(2,5),(10,1)$, if the ball passes directly through $D^{\prime}, E^{\prime}$ in a certain rectangle (without bouncing off the rectangle's sides), then $k_{i j}=l_{i j}=k_{D^{\prime} \xi^{\prime}}= \pm \frac{1}{2}$.

In this case, by equation (1), $k_{i j}=\frac{1}{2}$, and $i=j$. But when $i=j$, $k_{D^{\prime} E^{\prime}}=-\frac{1}{2}$, which is a contradiction.

If the ball bounces only once between $D$ and $E$, then the ball passes through $D^{\prime}, E^{\prime}$ in two adjacent rectangles, and $k_{D^{\prime} E^{\prime}}= \pm 2$, $\pm 1, \pm \frac{1}{4}, \pm \frac{1}{8}$, but by equation (1), $k_{i j} \neq k_{D^{\prime} E^{\prime}}$, which is a contradiction.

Therefore, the ball must bounce off the billiard table's sides at least twice between $D$ and $E$, and
$$
k_{02}=\frac{5}{2}=l_{00}(2,5)=l_{04}(10,1) .
$$

Thus, $f(2,5)=f(10,1)=2 \sqrt{29}$.
In summary, there are 11 integer points $E(a, b)$, and $f(E)_{\max }=2 \sqrt{37}$.