(11) As shown in the figure, let $S-ABCD$ be a pyramid with a height of 3, where the base $ABCD$ is a square with side length 2, and the projection of vertex $S$ onto the base is the center of the square $ABCD$. $K$ is the midpoint of edge $SC$, and a plane through $AK$ intersects segments $SB$ and $SD$ at points $M$ and $N$ respectively. ($M$ and $N$ can be the endpoints of the segments).
(1) Find the sine of the angle between line $AK$ and plane $SBC$;
(2) Prove that $\frac{V_{S-MKN}}{V_{S-BCD}}=\frac{SM \cdot SK \cdot SN}{SB \cdot SC \cdot SD}$, and find the volume of the pyramid $S-AMKN$ when $M$ is the midpoint of $SB$.
(11 (1) Method 1 Let the angle between $A K$ and the plane $S B C$ be $\theta$. Since
$$
S C=\sqrt{3^{2}+(\sqrt{2})^{2}}=\sqrt{11},
$$

we have $C K=\frac{\sqrt{11}}{2}$. Therefore,
$$
\cos \angle S C A=\frac{\sqrt{2}}{\sqrt{11}}=\frac{\sqrt{22}}{11} .
$$

Thus,
$$
A K^{2}=A C^{2}+C K^{2}-2 A C \cdot C K \cdot \cos \angle S C A=\frac{27}{4} .
$$

So $A K=\frac{3 \sqrt{3}}{2}$. Since
$$
V_{S-A B C}=\frac{1}{3} \times \frac{1}{2} \times 2 \times 2 \times 3=2=V_{A-S C B},
$$

we have $h_{A}=\frac{6}{S_{\triangle S B C}}=\frac{3 \sqrt{10}}{5}$. Therefore,
$$
\sin \theta=\frac{h_{A}}{A K}=\frac{2 \sqrt{30}}{15} .
$$

Method 2 Let $A C \cap B D=O$, and establish a spatial rectangular coordinate system with $O$ as the origin, $O A$ as the $x$-axis, $O B$ as the $y$-axis, and $O S$ as the $z$-axis. Then, $A(\sqrt{2}, 0,0), B(0, \sqrt{2}, 0), C(-\sqrt{2}, 0,0), S(0,0,3)$. Therefore,
$$
\overrightarrow{S B}=(0, \sqrt{2},-3), \overrightarrow{S C}=(-\sqrt{2}, 0,-3), K\left(-\frac{\sqrt{2}}{2}, 0, \frac{3}{2}\right) \text {. }
$$

Let $\vec{m}$ be a normal vector of the plane $S B C$, and it is easy to find that $\vec{m}=\left(-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}, 1\right)$.
Let $\theta$ be the angle between $A K$ and the plane $S B C$, since $\overrightarrow{A K}=\left(-\frac{3 \sqrt{2}}{2}, 0, \frac{3}{2}\right)$, we have
$$
\sin \theta=\left|\frac{\vec{m} \cdot \overrightarrow{A K}}{|\vec{m}| \cdot|\overrightarrow{A K}|}\right|=\frac{2 \sqrt{30}}{15} .
$$
(2) Let the projections of $N$ and $D$ on the plane $S B C$ be $H$ and $H_{1}$, respectively, then $\frac{N H}{D H_{1}}=$ $\frac{S N}{S D}$, and
$$
\frac{S_{\triangle S M K}}{S_{\triangle S B C}}=\frac{S M \cdot S K}{S B \cdot S C},
$$

so
$$
\frac{V_{S-M K N}}{V_{S-B C D}}=\frac{\frac{1}{3} \cdot N H \cdot S_{\triangle S M K}}{\frac{1}{3} \cdot D H_{1} \cdot S_{\triangle S B C}}=\frac{S M \cdot S K \cdot S N}{S B \cdot S C \cdot S D} .
$$

When $M$ is the midpoint of $S B$, $M K / / B C$.
Since $B C / /$ the plane $S A D$, we have $M K / /$ the plane $S A D$, so $M K / / A N$, $M K / / A D$, so $N$ and $D$ coincide.
It is easy to find that the distance from point $M$ to the plane $S A K$ is $\frac{\sqrt{2}}{2}$,
$$
S_{\triangle S A K}=\frac{1}{2} S_{\triangle S A C}=\frac{3 \sqrt{2}}{2} .
$$

Therefore,
$$
\begin{aligned}
V_{S-A M K N} & =\frac{1}{3} \cdot S_{\triangle S A K} \cdot \sqrt{2}+\frac{1}{3} \cdot S_{\triangle S A K} \cdot \frac{\sqrt{2}}{2} \\
& =\frac{\sqrt{2}}{2} \cdot S_{\triangle S A K}=\frac{3}{2} .
\end{aligned}
$$