1) In $\triangle ABC$, the lengths of the three sides are $a, b, c$, with $b < c$, and $AD$ is the angle bisector of $\angle A$, with point $D$ on $BC$.
(1) Find the necessary and sufficient condition for the existence of points $E$ and $F$ (not endpoints) on segments $AB$ and $AC$ respectively, such that $BE = CF$ and $\angle BDE = \angle CDF$ (expressed using $\angle A$, $\angle B$, $\angle C$);
(2) Given the existence of points $E$ and $F$, express the length of $BE$ in terms of $a, b, c$.
(1) As shown in Figure 1, if there exist points $E, F$ on segments $A B, A C$ such that $B E = C F$ and $\angle B D E = \angle C D F$, then because point $D$ is equidistant from $A B$ and $A C$, we have
$$
S_{\triangle B D E} = S_{\triangle C D F}
$$

Thus, we have
$$
B D \cdot D E = D C \cdot D F
$$

By the Law of Cosines, we have
$$
\begin{array}{l}
B D^{2} + D E^{2} - 2 B D \cdot D E \cos \angle B D E = B E^{2} = \\
C F^{2} = C D^{2} + D F^{2} - 2 C D \cdot D F \cos \angle C D F
\end{array}
$$

Since $\angle B D E = \angle C D F$, from (1) and (2), we can get
$$
\begin{aligned}
B D^{2} + D E^{2} & = C D^{2} + D F^{2} \\
B D + D E & = C D + D F
\end{aligned}
$$

From (1) and (4), one of the following two scenarios must hold:
$$
\begin{array}{l}
\left\{\begin{array}{l}
B D = C D \\
D E = D F
\end{array}\right. \\
\left\{\begin{array}{l}
B D = D F \\
D E = C D
\end{array}\right.
\end{array}
$$

Since $\angle D A F = \frac{1}{2} \angle A \\
2 \angle B > \angle A
\end{array}
$$

This is the necessary condition for the existence of points $E$ and $F$.
Conversely, if $2 \angle B > \angle A$, then
$$
\begin{array}{l}
\angle A D C = \angle B + \frac{1}{2} \angle A > \angle A \\
\angle A D B = \angle C + \frac{1}{2} \angle A > \angle A
\end{array}
$$

Thus, we can construct $\angle B D E = \angle A, \angle C D F = \angle A$ inside $\triangle A B C$, with point $E$ on $A B$ and point $F$ on $A C$. This leads to $A, C, D, E$ and $A, B, D, F$ being cyclic quadrilaterals. Therefore,

Thus
$$
\begin{array}{c}
\angle B = \angle D F C \\
\triangle B D E \backsim \triangle F D C
\end{array}
$$

Since these two triangles have equal heights from point $D$, we have $\triangle B D E \cong \triangle F D C$. Hence,
$$
B E = C F
$$

In summary, the necessary and sufficient condition is $2 \angle B > \angle A$.
(2) From the discussion in (1), we know that $A, C, D, E$ and $A, B, D, F$ are cyclic quadrilaterals. Therefore,
$$
B E \cdot B A = B D \cdot B C, C F \cdot C A = C D \cdot C B
$$

Since $B E = C F$, we have
$$
\begin{array}{c}
B E(A B + A C) = B C(B D + D C) = B C^{2} \\
B E = \frac{a^{2}}{b + c}
\end{array}
$$