II. (20 points) Given the line $l: y=\sqrt{3} x+4$, the moving circle $\odot O: x^{2}+y^{2}=r^{2}(1<r<2)$, and the rhombus $A B C D$ with one interior angle of $60^{\circ}$, where vertices $A$ and $B$ lie on line $l$, and vertices $C$ and $D$ lie on $\odot O$. Find the range of the area $S$ of rhombus $A B C D$ as $r$ varies.
Because rhombus $ABCD$ has an interior angle of $60^{\circ}$, $\triangle ACD$ or $\triangle BCD$ is an equilateral triangle. Without loss of generality, let $\triangle ACD$ be the equilateral triangle, as shown in Figure 3.

Since the distance from the center $O$ to the line $l$ is $2 > r$, the line $l$ is separated from $\odot O$.
Let $l_{CD}: y = \sqrt{3} x + b$.
Then the distance $d$ from the line $l$ to $CD$ is $d = \frac{|b-4|}{2}$.
Also, the distance from the center $O$ to the line $CD$ is $\frac{|b|}{2}$, so
$|CD| = 2 \sqrt{r^{2} - \left(\frac{|b|}{2}\right)^{2}} = \sqrt{4 r^{2} - b^{2}}$.
By $\left.d = \frac{\sqrt{3}}{2} \right| CD |$
$\Rightarrow \frac{|b-4|}{2} = \frac{\sqrt{3}}{2} \sqrt{4 r^{2} - b^{2}}$
$\Rightarrow b^{2} - 2b + 4 = 3 r^{2}$.
Since $1 < r < 2$, we have
$$
\begin{array}{l}
3 < b^{2} - 2b + 4 < 12 \\
\Rightarrow -2 < b < 1 \text{ or } 1 < b < 4. \\
\text{Also, } S = 2 S_{\triangle ACD} = 2 \times \frac{\sqrt{3}}{4} |CD|^{2} \\
= 2 \times \frac{\sqrt{3}}{4} \left(\frac{2}{\sqrt{3}}\right)^{2} d^{2} = \frac{\sqrt{3}}{6} (b-4)^{2},
\end{array}
$$

The function $S$ is monotonically decreasing in the intervals $(-2,1)$ and $(1,4)$, respectively. Therefore, the range of the area $S$ of rhombus $ABCD$ is $\left(0, \frac{3 \sqrt{3}}{2}\right) \cup \left(\frac{3 \sqrt{3}}{2}, 6 \sqrt{3}\right)$.