解: （1）如图，连接$BO，$延长$AO$交$\odot O$于点$D。$因为$PA，$$PB$是$\odot O$的切线，所以$PA\perp OA，$$PB\perp OB，$所以$\angle OAP = \angle OBP = 90^{\circ}，$所以$\angle AOB+\angle P = 360^{\circ}-\angle OAP-\angle OBP = 180^{\circ}。$因为$\angle AOB+\angle BOD = 180^{\circ}，$所以$\angle BOD = \angle P。$因为$\angle AOC = \angle P，$所以$\angle AOC = \angle BOD，$所以$\angle COB + 2\angle AOC = 180^{\circ}。$因为$OC = OB，$所以$\angle CBO=\angle BCO，$所以$\angle COB + 2\angle BCO = 180^{\circ}，$所以$\angle AOC = \angle BCO，$所以$BC// AO。$
（2）如图，延长$BC$交$PA$于点$E，$过点$O$作$OF\perp BC$于点$F，$则$\angle OFC = 90^{\circ}。$因为$BC = 10，$$CO = AO = BO = 13，$所以$BF = CF=\frac{1}{2}BC = 5，$所以$FO=\sqrt{CO^{2}-CF^{2}} = 12。$由（1），得$\angle OAP = 90^{\circ}，$$BC// AO，$所以$\angle AEF = 180^{\circ}-\angle OAP = 90^{\circ}，$所以四边形$AOFE$是矩形，所以$AE = FO = 12，$$EF = AO = 13。$因为$PA，$$PB$是$\odot O$的切线，所以$PA = PB。$设$PA = PB = x，$则$PE = PA - AE = x - 12。$因为$\angle AEF = 90^{\circ}，$所以$\angle PEB = 180^{\circ}-\angle AEF = 90^{\circ}。$因为$BE = BF + EF = 18，$$PB^{2}=PE^{2}+BE^{2}，$所以$x^{2}=(x - 12)^{2}+18^{2}，$解得$x=\frac{39}{2}，$所以$PA$的长为$\frac{39}{2}。$
