II. (25 points) As shown in Figure 3, in trapezoid $ABCD$, $AB \parallel CD$, $AD=12$, and $E$ is a point on side $CD$ such that $\frac{CE}{ED} = \frac{5}{4}$. Let the radius of the circle $\odot O_{1}$ passing through points $A, B, C, E$ be $R_{1}$, and the radius of the circle $\odot O_{2}$ passing through points $A, C, D$ be $R_{2}$. The side $BC$ is tangent to $\odot O_{2}$.
(1) Find the length of side $CD$;
(2) Find the range of $\frac{R_{1}}{R_{2}}$.
(1) Since $BC$ is the tangent of $\odot O_{2}$, therefore, $\angle ACB = \angle CDA$.
Also, $AB // CD$, so $\angle BAC = \angle ACD$.
Thus, $\triangle ABC \backsim \triangle CAD$.
Hence, $\angle ABC = \angle CAD$.
Therefore, $AD$ is the tangent of $\odot O_{1}$.
By the secant-tangent theorem, we have $AD^2 = DE \cdot DC$.
Thus, $12^2 = \frac{4}{9} DC^2$. Solving this, we get $CD = 18$.
(2) From (1), we know that $\triangle ABC \backsim \triangle CAD$, and $CD = 18$.

Therefore, $\frac{R_{1}}{R_{2}} = \frac{AC}{CD}$, which means $\frac{R_{1}}{R_{2}} = \frac{AC}{18}$.
Since $6 = CD - AD < AC < CD + AD = 30$, we have
$$
\frac{1}{3} < \frac{R_{1}}{R_{2}} < \frac{5}{3}.
$$

Also, $AC \neq CD$ (otherwise, quadrilateral $ABCD$ would be a parallelogram), so $\frac{R_{1}}{R_{2}} \neq 1$.
Hence, the range of $\frac{R_{1}}{R_{2}}$ is $\frac{1}{3} < \frac{R_{1}}{R_{2}} < \frac{5}{3}$, and $\frac{R_{1}}{R_{2}} \neq 1$.