（1）$\angle ABC$与$\angle C$相等，理由如下：
连接$OD，$
$\because FD$切$\odot O$于点$D，$
$\therefore OD\perp DF，$
$\because BF\perp DF，$
$\therefore OD// BF，$
$\therefore \angle ODB=\angle FBD，$
$\because OB=OD，$
$\therefore \angle ODB=\angle OBD，$
$\therefore \angle OBD=\angle FBD，$
$\because AC// BF，$
$\therefore \angle C=\angle FBD，$
$\therefore \angle C=\angle OBD，$
$\therefore \angle ABC=\angle C；$
（2）连接$BE，$$OE，$
$\because \widehat{DG}=60^\circ，$
$\therefore \angle GOD=60^\circ，$
$\therefore \angle FBD=\frac{1}{2}\angle GOD=30^\circ，$
$\because AC// BF，$
$\therefore \angle C=\angle FBD=30^\circ，$
由（1）结论$\angle ABC=\angle C$得$\angle ABC=30^\circ，$
$\because \angle BAE=\angle C+\angle ABC，$
$\therefore \angle BAE=60^\circ，$
$\because OA=OE，$
$\therefore \triangle OAE$是等边三角形，
$\therefore \angle EOA=60^\circ，$
$\therefore \angle ABE=\frac{1}{2}\angle EOA=30^\circ，$
$\therefore \angle ABE=\angle ABC，$
$\therefore \widehat{AD}=\widehat{AE}，$
$\therefore$点$D$与点$E$关于直线$AB$对称.
