1.连AC∵∠ABC=∠BAD=90°AB=BC=AD/2∴CD⊥AC∵PA⊥面ABCD∴PA⊥CD∴CD⊥面PAC又CD∈面PCD∴面PCD⊥面PAC2.延长DA至F,使AF=AB,连PF则∠BPF为异面直线AE与PB所成的角，设为α设AB=1则BF=√2，PB=PF=2cosα=(PB²+PF²-BF²)/(2·PB·PF)=6/8=3/4
