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$(3,-2)$

$30^{\circ}$

$35^{\circ}$

 解：（1）$\because \triangle BAD\cong\triangle ACE，$

 $\therefore BD = AE，$$AD = CE。$

 又$\because A，$$D，$$E$三点在同一条直线上，

 $\therefore AE = DE + AD。$

 $\therefore BD = DE + CE$

 （2）$\because \triangle BAD\cong\triangle ACE，$

 $\therefore AB = CA，$$\angle BAD=\angle ACE。$

 $\because \angle E = 90^{\circ}，$

 $\therefore \angle CAE+\angle ACE = 90^{\circ}。$

 $\therefore \angle CAE+\angle BAD = 90^{\circ}，$即$\angle BAC = 90^{\circ}。$

 $\therefore \triangle ABC$是等腰直角三角形

 （3）答案不唯一，如将$\triangle BAD$先绕点$D$按顺时针方向旋转$90^{\circ}，$再向下平移，即可与$\triangle ACE$完全重合

 解：（1）$\because \triangle CAD\cong\triangle CED，$$\triangle CEF\cong\triangle CAD，$

 $\therefore \angle ACD=\angle ECD，$$\angle ECF=\angle ACD。$

 $\therefore \angle ACD=\angle ECD=\angle ECF。$

 $\because \angle ACD+\angle ECD+\angle ECF=\angle ACB = 90^{\circ}，$

 $\therefore \angle ACD=\angle ECD=\angle ECF = 30^{\circ}。$

 又$\because \triangle CEF\cong\triangle BEF，$

 $\therefore \angle B=\angle ECF = 30^{\circ}。$

 $\therefore \angle A = 90^{\circ}-\angle B = 60^{\circ}$

 （2）$AC// EF$

 理由：$\because \triangle CEF\cong\triangle BEF，$

 $\therefore \angle CFE=\angle BFE。$

 又$\because \angle BFE+\angle CFE = 180^{\circ}，$

 $\therefore \angle BFE = 90^{\circ}。$

 $\therefore \angle ACB=\angle BFE。$

 $\therefore AC// EF$