解: （1）四边形$AFHE$是正方形
理由：$\because$ $Rt\triangle ABE$绕点$A$按逆时针方向旋转$90^{\circ}$得到$\triangle ADF，$$\therefore$ $Rt\triangle ABE\cong Rt\triangle ADF。$
$\therefore$ $\angle AEB=\angle AFD = 90^{\circ}，$$\therefore$ $\angle AFH = 90^{\circ}。$$\because$ $Rt\triangle ABE\cong Rt\triangle ADF，$$\therefore$ $AE = AF，$$\angle BAE=\angle DAF。$
又$\because$ 易得$\angle DAF+\angle FAB = 90^{\circ}，$$\therefore$ $\angle BAE+\angle FAB = 90^{\circ}，$$\therefore$ $\angle FAE = 90^{\circ}。$
在四边形$AFHE$中，$\angle FAE = 90^{\circ}，$$\angle AEH = 90^{\circ}，$$\angle AFH = 90^{\circ}，$
$\therefore$ 四边形$AFHE$是矩形。又$\because$ $AE = AF，$$\therefore$ 四边形$AFHE$是正方形。
（2）设$AE = x。$易得$AE = EH = FH = x，$$BC = AB = 13。$
在$Rt\triangle AEB$中，$AB^{2}=AE^{2}+BE^{2}，$即$13^{2}=x^{2}+(x + 7)^{2}，$
解得$x_{1}=5，$$x_{2}=-12$（不合题意，舍去）。$\therefore$ $EH = FH = 5。$
$\therefore$ $BE = BH+EH = 7 + 5 = 12。$$\because$ $Rt\triangle ABE\cong Rt\triangle ADF，$
$\therefore$ $DF = BE = 12。$$\therefore$ $DH = DF+FH = 12 + 5 = 17$
