解：（1）连接$PD，$$PC。$
因为$OA$、$OB$都是$\odot P$的切线，所以$\angle ODP = \angle OCP = 90^\circ。$
已知$\angle AOB = 60^\circ，$四边形内角和为$360^\circ，$则$\angle DPC = 360^\circ - 90^\circ - 90^\circ - 60^\circ = 120^\circ。$
$\odot P$的半径为$3\,\text{cm}，$劣弧$CD$的长为$\frac{120\pi \times 3}{180} = 2\pi\,\text{cm}。$
（2）分两种情况：
① 点$P$在$\angle AOB$内，连接$PE$、$PC，$延长$CP$交直线$OB$于点$N，$过点$P$作$PM \perp EF，$垂足为$M。$
因为$EF = 4\sqrt{2}\,\text{cm}，$所以$EM = \frac{1}{2}EF = 2\sqrt{2}\,\text{cm}。$
在$Rt\triangle EPM$中，$PE = 3\,\text{cm}，$$EM = 2\sqrt{2}\,\text{cm}，$则$PM = \sqrt{3^2 - (2\sqrt{2})^2} = 1\,\text{cm}。$
因为$\angle AOB = 60^\circ，$$\angle OCP = 90^\circ，$所以$\angle ONC = 30^\circ。$
在$Rt\triangle MNP$中，$\angle ONC = 30^\circ，$$PM = 1\,\text{cm}，$则$PN = 2PM = 2\,\text{cm}。$
所以$CN = CP + PN = 3 + 2 = 5\,\text{cm}。$
在$Rt\triangle OCN$中，$\angle ONC = 30^\circ，$$CN = 5\,\text{cm}，$则$OC = \frac{CN}{\sqrt{3}} = \frac{5\sqrt{3}}{3}\,\text{cm}。$
② 点$P$在$\angle AOB$外，连接$PF$、$PC，$$PC$交$EF$于点$N，$过点$P$作$PM \perp EF，$垂足为$M。$
同理可得$PM = 1\,\text{cm}，$$PN = 2\,\text{cm}，$则$CN = CP - PN = 3 - 2 = 1\,\text{cm}。$
在$Rt\triangle OCN$中，$\angle ONC = 30^\circ，$$CN = 1\,\text{cm}，$则$OC = \frac{CN}{\sqrt{3}} = \frac{\sqrt{3}}{3}\,\text{cm}。$
综上，$OC$的长为$\frac{5\sqrt{3}}{3}\,\text{cm}$或$\frac{\sqrt{3}}{3}\,\text{cm}。$
答案：（1）$2\pi\,\text{cm}；$（2）$\frac{5\sqrt{3}}{3}\,\text{cm}$或$\frac{\sqrt{3}}{3}\,\text{cm}。$
