(1)连接B1C,D1C
由题意得该棱柱为正四棱柱,且BC=CD=2,CE=1,BB1=DD1=4
易证△B1BC∽△BCE,由相似得B1C⊥BE
∵A1B1⊥面BB1C1C,∴A1C在面BB1C1C的射影是B1C
∴A1C⊥BE
同理A1C⊥DE,∴A1C⊥面BDE
(2)设A1C与面BDE交於O,连接OE,则A1O⊥OE
勾股定理得A1E=√17,A1C=√24
在△A1CE中,馀弦定理得cosCA1E=10/√102
∴sinA1EO=cosCA1E=10/√102,即A1E与面BDE所成角的正弦值为10/√102
同理,在△A1DE中,馀弦定理得cosA1ED=1/√85,∴sinA1ED=√84/√85,即A1E与二面角的棱所成角的正弦值为√84/√85
设二面角为θ,由三正弦定理得sinθ=sinA1EO/sinA1ED=5√85/√2142
∴cosθ=1/√126=√14/42