21. Given that the vertex of parabola $G$ is at the origin, and its focus is on the positive $y$-axis, the distance from point $P(m, 4)$ to its directrix is 5.
(1) Find the equation of parabola $G$;
(2) As shown in the figure, a line passing through the focus of parabola $G$ intersects the parabola $G$ and the circle $x^{2}+(y-1)^{2}=1$ at points $A, C, D, B$ respectively. Prove that $|A C| \cdot|B D|$ is a constant;
(3) Draw the tangents $l_{1}, l_{2}$ to parabola $G$ at points $A, B$ respectively, and let $l_{1}, l_{2}$ intersect at point $M$. Find the minimum value of the sum of the areas of $\triangle A C M$ and $\triangle B D M$.
(1) $\frac{p}{2}=1 \Rightarrow p=2$, so the equation of the parabola $G$ is $x^{2}=4 y$.
(2) Let the line $A B: y=k x+1, A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
$C\left(x_{3}, y_{3}\right), D\left(x_{4}, y_{4}\right)$, combining with $x^{2}=4 y$ we get
$$
x^{2}-4 k x-4=0 \text {, }
$$

Combining with $x^{2}+(y-1)^{2}=1$ we get $\left(k^{2}+1\right) x^{2}-1=0$.
Since $x_{3}+x_{4}=0$, then
$$
\begin{aligned}
& \left(x_{3}-x_{1}\right)\left(x_{2}-x_{4}\right)=x_{1} x_{4}+x_{2} x_{3}-x_{1} x_{2}-x_{3} x_{4} \\
= & -x_{4}\left(x_{2}-x_{1}\right)+4-x_{3} x_{4}=-\frac{1}{\sqrt{k^{2}+1}} \cdot 4 \sqrt{k^{2}+1}+4+\frac{1}{k^{2}+1}=\frac{1}{k^{2}+1},
\end{aligned}
$$

Therefore, $|A C| \cdot|B D|=\sqrt{k^{2}+1}\left|x_{3}-x_{1}\right| \cdot \sqrt{k^{2}+1}\left|x_{2}-x_{4}\right|=1$ is a constant.
(3) The equations of the tangents at points $A, B$ are $\left\{\begin{array}{l}x_{1} x=2\left(y+y_{1}\right), \\ x_{2} x=2\left(y+y_{2}\right)\end{array} \Rightarrow\left\{\begin{array}{l}x=\frac{x_{1}+x_{2}}{2}=2 k, \\ y=\frac{x_{1} x_{2}}{4}=-1,\end{array}\right.\right.$

Thus, $M(2 k,-1)$. Then $k_{M F}=-\frac{1}{k} \Rightarrow M F \perp A B$, and $|M F|=2 \sqrt{k^{2}+1}$.
Also, $|A C|+|B D|=\sqrt{k^{2}+1}\left(\left|x_{3}-x_{1}\right|+\left|x_{2}-x_{4}\right|\right)=\sqrt{k^{2}+1}\left[\left(x_{2}-x_{1}\right)-\left(x_{4}-x_{3}\right)\right]$ $=\sqrt{k^{2}+1}\left(4 \sqrt{k^{2}+1}-\frac{2}{\sqrt{k^{2}+1}}\right)=4\left(k^{2}+1\right)-2$,
Thus, $S_{\triangle A C M}+S_{\triangle B D M}=4\left(\sqrt{k^{2}+1}\right)^{3}-2 \sqrt{k^{2}+1}$, let $t=\sqrt{k^{2}+1} \geqslant 1$, $f(t)=4 t^{3}-2 t$, then $f^{\prime}(t)=12 t^{2}-2>0 \Rightarrow f(t)$ is monotonically increasing, so when $k=0$, $S_{\triangle A C M}+S_{\triangle B D M}$ reaches its minimum value of 2.