$\ 解：（1）如图①，连接AH$
$因为线段EA绕点E顺时针旋转得到EH，点H落在边CD上，所以EA = EH$
$因为四边形ABCD是正方形，所以CB = CD = AB = AD，\angle ABE = \angle ADH = 90^{\circ}$
$又因为CE = CH，所以CB - CE = CD - CH，即BE = DH$
$在\triangle ABE和\triangle ADH中,\begin{cases}AB = AD\\\angle ABE = \angle ADH\\BE = DH\end{cases}$
$所以\triangle ABE\cong\triangle ADH（SAS）$
$所以AE = AH。\ 又因为EA = EH，所以AE = AH = EH$
$所以\triangle AEH为等边三角形，所以\angle AEH = 60^{\circ}$
$(2）如图②，在AB上取点Q，使得BQ = BE，过点H作HI\perp BF于点I$
$因为线段EA绕点E顺时针旋转得到EH，点H落在边CG上，所以EA = EH$
$因为AB = BC，BQ = BE，所以AQ = CE$
$因为CG平分\angle DCF，所以CI = HI$
$设AQ = CE = a，CI = HI = b，BQ = BE = x$
$在Rt\triangle ABE中，AE^2 = AB^2 + BE^2，在Rt\triangle EIH中，EH^2 = EI^2 + IH^2$
$所以AB^2 + BE^2 = EI^2 + IH^2，即(a + x)^2 + x^2 = (a + b)^2 + b^2，整理得(x - b)(a + b + x)=0$
$因为a + b + x\neq0，所以x - b = 0，解得x = b，所以CI = BE$
$因为CG平分\angle DCF，BQ = BE，所以\angle GCF = \angle BQE = 45^{\circ}，所以\angle AQE = \angle ECH = 135^{\circ}$
$因为QE = \sqrt{BE^2 + BQ^2}=\sqrt{2}BE，CH = \sqrt{CI^2 + CH^2}=\sqrt{2}CI，所以QE = CH$
$在\triangle AQE和\triangle ECH中，\ \begin{cases}AQ = EC\\\angle AQE = \angle ECH\\QE = CH\end{cases}$
$所以\triangle AQE\cong\triangle ECH（SAS），所以\angle QAE = \angle CEH$
$因为\angle QAE+\angle AEB = 90^{\circ}，所以\angle CEH+\angle AEB = 90^{\circ}，所以\angle AEH = 90^{\circ}，所以AE\perp EH$
$（3）HP^2 + AP^2 = 2EP^2。\ 理由：$
$如图③，过点P作PM\perp EH于点M，PN\perp AE于点N$
$由（2）可知，EA = EH，AE\perp EH，所以\angle EAH = \angle EHA = 45^{\circ}$
$所以\triangle APN和\triangle PHM为等腰直角三角形，所以HP = \sqrt{2}PM，AP = \sqrt{2}PN$
$所以PH^2 = 2PM^2，AP^2 = 2PN^2$
$因为\angle PNE = \angle AEH = \angle PME = 90^{\circ}，所以四边形PNEM为矩形，所以PM = NE$
$在Rt\triangle PNE中，EP^2 = NE^2 + PN^2 = PM^2 + PN^2，所以HP^2 + AP^2 = 2(PM^2 + PN^2)=2EP^2$
