As shown in Figure $3, D$ and $E$ are points on side $BC$ of $\triangle ABC$, and $F$ is a point on the extension of $BC$. $\angle DAE = \angle CAF$.
(1) Determine the positional relationship between the circumcircle of $\triangle ABD$ and the circumcircle of $\triangle AEC$, and prove your conclusion.
(2) If the radius of the circumcircle of $\triangle ABD$ is twice the radius of the circumcircle of $\triangle AEC$, and $BC=6, AB=4$, find the length of $BE$.
Three, (1) Two circles are externally tangent.
Draw the tangent line $l$ of $\odot A B D$, then $\angle 1=\angle B$.
$$
\begin{array}{l}
\because \angle 3=\angle B+\angle C, \\
\therefore \angle 3=\angle 1+\angle C . \\
\because \angle 1+\angle 2=\angle 3=\angle 1+\angle C, \\
\therefore \angle 2=\angle C .
\end{array}
$$

Draw $A P \perp l$ through $A$, intersecting $\odot A E C$ at point $P$, and connect $P E$.
$\because \angle P=\angle A C E$, then $\angle 2=\angle P$.
$$
\therefore \angle P A E+\angle P=90^{\circ} \text {. }
$$

Thus, $\angle A E P=90^{\circ}$, so $A P$ is the tangent line of $\odot A E C$, i.e., the circle is tangent at point $A$.
(2) Extend $D A$ to intersect $\odot A E C$ at $G$ (assuming $F$ is on $\odot A E C$), and connect $(S F$.
$$
\text { Since } \angle 4=\angle D A E+\angle A E D=\angle 3+\angle A F C \text {, }
$$

we have $\angle 4+\angle 5=180^{\circ}$, then $\angle 4=\angle A G F$.
$\therefore \triangle A D B \backsim \triangle A G F$.
$\therefore A B: A F=2$ (i.e., equal to the ratio of the radii of the two circles).
But $A B=4$,
$\therefore A F=2$. (This can also be done using the sine rule)
$\because B A \cdot B F=B E \cdot B C$,
$\therefore B E=4$.
(Provided by Yu Qiao)