II. (25 points) As shown in Figure 3, circles $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at point $O$. Line $AB$ is tangent to $\odot O_{1}$ and $\odot O_{2}$ at points $B$ and $A$, respectively, and intersects the $x$-axis and $y$-axis at points $M(2 \sqrt{3}, 0)$ and $C(0,2)$.
(1) Find the radius of $\odot O_{2}$;
(2) Does there exist a point $P$ on line $AB$ such that $\triangle M O_{2} P$ is similar to $\triangle M O B$? If so, find the coordinates of point $P$; if not, explain why.
(1) It is easy to know that $l_{A B}: y=-\frac{\sqrt{3}}{3} x+2$.
Then $\angle A M O_{2}=30^{\circ}$.
It is easy to know that $O$ is the midpoint of $M \mathrm{O}_{2}$.
Therefore, the radius of $\odot \mathrm{O}_{2}$ is $R=2 \sqrt{3}$.
(2) It is easy to see that $\triangle M O B$ is an isosceles triangle, and
$\angle B O M=30^{\circ}$.
If there exists a point $P$ that satisfies the condition, then
$\angle M O_{2} P=30^{\circ}$ or $\angle M P O_{2}=30^{\circ}$.
When $\angle M O_{2} P=30^{\circ}$, by $\angle M O_{2} C=30^{\circ}$, we know that point $P$ coincides with $C$. At this time, $P(0,2)$.

When $\angle M P O_{2}=30^{\circ}$, draw $P H \perp x$-axis at point $H$, then Rt $\triangle P O_{2} A \cong \mathrm{Rt} \triangle P O_{2} H$.
Thus, $O_{2} H=O_{2} A=2 \sqrt{3}$.
Therefore, $P(-4 \sqrt{3}, 6)$.
In summary, the solutions are $P(0,2)$ or $P(-4 \sqrt{3}, 6)$.