解：连接$OD，$$OE。$因为$DF$是$\odot O$的切线，所以$OD\perp DF，$所以$\angle ODF = 90^{\circ}。$因为$\angle CDF = 22.5^{\circ}，$所以$\angle ODB = 180^{\circ}-\angle ODF - \angle CDF = 67.5^{\circ}。$因为$OB = OD，$所以$\angle B = \angle ODB = 67.5^{\circ}。$因为$AB = AC，$所以$\angle C = \angle B = 67.5^{\circ}，$所以$\angle BAC = 180^{\circ}-\angle B - \angle C = 45^{\circ}。$因为$OA = OE，$所以$\angle OEA = \angle BAC = 45^{\circ}，$所以$\angle AOE = 180^{\circ}-\angle BAC - \angle OEA = 90^{\circ}。$因为$\odot O$的半径为$4，$所以$OA = OE = 4，$所以$S_{\triangle OAE}=\frac{1}{2}OA\cdot OE = 8，$$S_{扇形OAE}=\frac{90\pi\times4^{2}}{360}=4\pi，$所以$S_{阴影}=S_{扇形OAE}-S_{\triangle OAE}=4\pi - 8。$
