答案：
解: (1) $\triangle A B D \cong \triangle C A E$.
证明: $\because \triangle A B C$ 为等边三角形,
$$
\therefore A B=A C, \angle B A C=\angle B=60^{\circ} \text {, }
$$
在 $\triangle A B D$ 和 $\triangle C A E$ 中,
$$
\begin{aligned}
& \left\{\begin{array}{l}
A B=C A \\
\angle A B D=\angle C A E, \\
B D=A E
\end{array}\right. \\
& \therefore \triangle A B D \cong \triangle C A E(S A S) ; \\
& \therefore \angle B A D=\angle A C E, \\
& \therefore \angle A F E=\angle A C E+\angle F A C=\angle B A D+\angle F A C=\angle B A C=60^{\circ} .
\end{aligned}
$$
(2)(1)中的结论是否仍然成立.
理由: $\because \triangle A B C$ 为等边三角形,
$$
\therefore A B=A C, \angle B A C=\angle A B C=60^{\circ} \text {, }
$$
在 $\triangle A B D$ 和 $\triangle C A E$ 中,
$$
\begin{aligned}
& \left\{\begin{array}{l}
A B=C A \\
\angle A B D=\angle C A E, \\
B D=A E
\end{array}\right. \\
& \therefore \triangle A B D \cong \triangle C A E(S A S) ; \\
& \therefore \angle D=\angle E, \\
& \because \angle A B C=\angle E+\angle B C E, \angle A F E=\angle D+\angle D C F, \angle B C E=\angle D C F, \\
& \therefore \angle A F E=\angle E+\angle B C E=\angle A B C=60^{\circ} .
\end{aligned}
$$