(1)证明:∵圆锥PO∴PO⊥底面圆O,∴PO⊥AC∵D为AC中点,∴OD⊥AC∴枣困AC⊥面POD(2)解析:∵∠CAB=30°,PO=√2,AB=2∴OD=1/2,AC=√3,PD=3/2由(1)易知面PAC⊥POD过O作OE⊥PD交PD于E设DE=x∵⊿POD为Rt⊿∴OD^2=ED*PD==>1/4=3/2*x==>x=1/蚂岩汪6∴OE=√(OD^2-DE^2)= √2/3∴直线OC与面PAC所成角的正闷仔弦值为√2/3