(1)联结AC.在菱形ABCD中,∵AB=BC,∠B=60°,∴△ABC是等边三角形.∴AC=AB,∠BAC=∠BCA=60°.∵∠PAQ=60°,∴∠BAP=∠CAQ.∵AB∥CD,∠B=60°,∴∠BCD=120°.∴∠ACQ=∠B=60°.∴△ABP≌△ACQ.∴AP=AQ.∴△APQ是等边三角形.(2)由△APQ是等边三角形,得AP=PQ=y.作AH⊥BC于点H,由AB=4,BH=2,∠B=60°,得AH=.∴,即.定义域为x≥0(3)(i)当点P在边BC上时,∵PD⊥AQ,AP=PQ,∴PD垂直平分AQ.∴AD=DQ.∴CQ=0.又∵BP=CQ,∴BP=0.(ii)当点P在边BC的延长线上时,同理可得BP=8.综上所述,BP=0或BP=8.