分析:(Ⅰ)欲证BF∥平面A'DE,只需在平面A'DE中找到一条线平行于BF即可;而取A′D的中点G,并连接GF、GE,易证四边形BEGF为平行四边形,则BF∥EG,即问题得证.
(Ⅱ)欲求直线FM与平面A′DE所成角的余弦值,需先找到直线FM与平面A′DE所成的角;而连接A′M,CE,由平面A′DE⊥平面BCD易证CE⊥A′M,且由勾股定理的逆定理可证CE⊥DE;再取A′E的中点N,连线NM、NF,则NF⊥平面A′DE,即∠FMN为直线FM与平面A′DE所成的角;最后在Rt△FMN中,易得cos∠FMN的值.
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