Example 3 As shown in Figure 2, the vertices of a large hexagram (thick solid line) are the centers of six congruent shaded small hexagrams around it, and each pair of adjacent small hexagrams share one vertex.

If the distance from vertex $C$ to the center $A$ of a small hexagram is $a$, find:
(1) the distance from vertex $A$ of the large hexagram to its center $O$;
(2) the area of the large hexagram;
(3) the ratio of the area of the large hexagram to the sum of the areas of the six small hexagrams.

保留源文本的换行和格式，直接输出翻译结果。
(1) As shown in Figure 2, connect $C O$. It is easy to see that
$$
\angle A C O=90^{\circ}, \angle A O C=30^{\circ} \text {. }
$$

Therefore, $A O=2 A C=2 a$.
(2) Observing Figure 2, we know that the area of the large hexagram is 12 times the area of the equilateral $\triangle A M N$.
From $A M^{2}=\left(\frac{A M}{2}\right)^{2}+\left(\frac{2 a}{2}\right)^{2}$, we solve to get
$$
A M=\frac{2 \sqrt{3}}{3} a \text {. }
$$

Therefore, the area of the large hexagram is
$$
S=12 \times \frac{1}{2} \times \frac{2 \sqrt{3}}{3} a \times a=4 \sqrt{3} a^{2} \text {. }
$$
(3) According to the given conditions, the distance from the vertex $A$ of the large hexagram to its center $O$ is $2 a$. Therefore, the area of the large hexagram is 4 times the area of a small hexagram. Thus,