II. (Full marks 25 points) As shown in the figure, $\odot O_{1}$ and $\odot O_{2}$ are externally tangent at $M$, and the angle between their two external common tangents is $60^{\circ}$. The line connecting the centers intersects $\odot O_{1}$ and $\odot O_{2}$ at $A$ and $B$ (different from $M$), respectively. A line through $B$ intersects $\odot O_{1}$ at points $C$ and $D$. Find the value of $\operatorname{ctg} \angle B A C \cdot \operatorname{ctg} \angle B A D$.
From the given conditions, we have $O_{1} E \perp P E, O_{1} F \perp P F, O_{1} E= O_{1} F$, thus $O_{1}$ lies on the bisector of $\angle E P F$. Similarly, $O_{2}$ lies on the bisector of $\angle E P F$. Therefore, $P A$ is the bisector of $\angle E P F$. Since $O_{2} Q / / P E$, we have $\angle Q O_{2} O_{1}=\angle E P O_{1}=30^{\circ}$, hence $O_{1} Q=\frac{1}{2} O_{1} O_{2}$.

Let the radii of the larger and smaller circles be $R$ and $r$. Then $O_{1} Q=R-r, O_{1} O_{2}=R+r$, so $R-r=\frac{1}{2}(R+r)$, thus $R=3 r$.
$\because \angle A C M=\angle A D M=90^{\circ}$,
$\therefore \operatorname{ctg} \angle B A C \cdot \operatorname{ctg} \angle B A D=\frac{A C}{C M} \cdot \frac{A D}{D M}$.
From $\triangle B A C \sim \triangle B D M, \triangle B A D \sim \triangle B C M$ we get
$$
\frac{A C}{D M}=\frac{A B}{D B}, \frac{A D}{C M}=\frac{D B}{M B} \text {. }
$$

Therefore, $\operatorname{ctg} \angle B A C \cdot \operatorname{ctg} \angle B A D=\frac{A B}{D B} \cdot \frac{D B}{M B}=\frac{8 r}{2 r}=4$.