证明:(1)如图, 连结AC,AD1,CD1,A1C1,A1B,C1B. ∵ABCD-A1B1C1D1是正方体,∴AA1∥CC1,AA1=CC1, ∴四边形AA1C1C为平行四边形,∴A1C1∥AC. A1C1⊄平面ACD1,AC⊂平面ACD1,∴A1C1∥平面ACD1; ∵A1D1∥BC,A1D1=BC,∴四边形A1BCD1为平行四边形,∴A1B∥CD1. A1B⊄平面ACD1,CD1⊂平面ACD1,∴A1B∥⊄平面ACD1, 又A1B∩A1C1=A1, ∴平面A1BC1∥平面ACD1; (2)连结C1F,∵E,F分别是棱AA1,BB1的中点,∴EF∥C1D1,EF=C1D1 ∴EFC1D1是平行四边形,∴D1F∥C1E. 设正方体ABCD-A1B1C1D1的棱长为2,解直角三角形求得A1C1=2,A1F=C1F=. 在△A1C1F中,由余弦定理得cos∠A1FC1===. ∴异面直线A1F与D1E所成的角的余弦值是.
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