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解: (1) 如图，连接$OC，$$OD。$因为$M$是$CD$的中点，$CD = 12，$所以$CM=\frac{1}{2}CD = 6。$因为$OC = OD，$所以$OM\perp CD，$所以$\angle OMC = 90^{\circ}。$因为$OM = 3，$所以$OC=\sqrt{OM^{2}+CM^{2}} = 3\sqrt{5}。$故$\odot O$的半径为$3\sqrt{5}。$

证明: (2) 如图，连接$AC，$延长$AF$交$BD$于点$G。$因为$AB\perp CD，$$CE = EF，$所以$\angle AEF = 90^{\circ}，$$AB$垂直平分$CF，$所以$AC = AF，$所以$\angle CAE=\angle FAE。$因为$\angle BDC=\angle CAE，$所以$\angle BDC=\angle FAE。$因为$\angle BDC+\angle DFG+\angle DGF = 180^{\circ}，$$\angle FAE+\angle AFE+\angle AEF = 180^{\circ}，$$\angle DFG=\angle AFE，$所以$\angle DGF=\angle AEF = 90^{\circ}，$所以$AF\perp BD。$

证明: (1) 因为$AB$是$\odot O$的直径，所以$\angle ACB = 90^{\circ}，$所以$\angle OCB+\angle ACO = 90^{\circ}。$因为$CP$是$\odot O$的切线，所以$OC\perp CP，$所以$\angle OCP = 90^{\circ}，$所以$\angle OCB+\angle BCP = 90^{\circ}，$所以$\angle ACO=\angle BCP。$

 (2) 因为$OA = OC，$所以$\angle OAC=\angle ACO=\angle BCP。$因为$\angle ABC = 2\angle BCP，$所以$\angle ABC = 2\angle OAC。$因为$\angle ACB = 90^{\circ}，$所以$\angle ABC+\angle OAC = 90^{\circ}，$所以$3\angle OAC = 90^{\circ}，$所以$\angle OAC = 30^{\circ}，$所以$\angle BOC = 2\angle OAC = 60^{\circ}。$因为$\angle OCP = 90^{\circ}，$所以$\angle P = 90^{\circ}-\angle BOC = 30^{\circ}。$

 (3) 因为$\angle ACB = 90^{\circ}，$$\angle OAC = 30^{\circ}，$$AB = 4，$所以$BC=\frac{1}{2}AB = 2，$所以$AC=\sqrt{AB^{2}-BC^{2}} = 2\sqrt{3}，$所以$S_{\triangle ABC}=\frac{1}{2}AC\cdot BC = 2\sqrt{3}。$因为$OA=\frac{1}{2}AB = 2，$所以$S_{半圆}=\frac{1}{2}\pi\cdot OA^{2}=2\pi，$所以$S_{阴影}=S_{半圆}-S_{\triangle ABC}=2\pi - 2\sqrt{3}。$故阴影部分的面积为$2\pi - 2\sqrt{3}。$

证明: (1) 连接$OD。$因为$CD = AC，$所以$\angle A=\angle ADC。$因为$OB = OD，$所以$\angle B=\angle ODB。$因为$\angle ACB = 90^{\circ}，$所以$\angle A+\angle B = 90^{\circ}，$所以$\angle ADC+\angle ODB = 90^{\circ}，$所以$\angle ODC = 180^{\circ}-(\angle ADC+\angle ODB)=90^{\circ}，$所以$OD\perp CD。$因为$OD$为$\odot O$的半径，所以$CD$为$\odot O$的切线。

 (2) 因为$\angle ACB = 90^{\circ}，$$\angle A = 60^{\circ}，$所以$\angle B = 90^{\circ}-\angle A = 30^{\circ}。$因为$CD = AC，$所以$\triangle ACD$为等边三角形，所以$\angle ACD = 60^{\circ}，$$CD = AC = 2\sqrt{3}，$所以$\angle OCD=\angle ACB-\angle ACD = 30^{\circ}，$所以$OD=\frac{1}{2}OC。$设$OB = OD = x，$则$OC = 2x，$所以$CD=\sqrt{OC^{2}-OD^{2}}=\sqrt{3}x，$所以$\sqrt{3}x = 2\sqrt{3}，$解得$x = 2，$所以$OB = OD = 2。$因为$\angle COD = 2\angle B = 60^{\circ}，$所以$\angle BOD = 180^{\circ}-\angle COD = 120^{\circ}，$所以$\overset{\frown}{BD}$的长为$\frac{120\pi\times2}{180}=\frac{4\pi}{3}。$