解答:(1)证明:∵PB⊥底面ABC,且AC?底面ABC,∴AC⊥PB,
由∠BCA=90°,可得AC⊥CB,
又∵PB∩CB=B,∴AC⊥平面PBC,
∵BE?平面PBC,∴AC⊥BE,
∵PB=BC,E为PC中点,∴BE⊥PC,
∵AC∩PC=C,∴BE⊥平面PAC,
∵BE?平面BEF,∴平面PAC⊥平面BEF;
(2)解:取AF的中点G,AB的中点M,连接CG,CM,GM,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/3c6d55fbb2fb4316d6ab36fd23a4462308f7d385?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∵E为PC的中点,2PF=AF,∴EF∥CG,
∵CG?平面BEF,EF?平面BEF,
∴CG∥平面BEF.
同理可证:GM∥平面BEF,∵CG∩GM=G,∴平面CMG∥平面BEF.
则平面CMG与平面平面BEF所成的二面角的平面角(锐角)就等于平面ABC与平面BEF所成的二面角的平面角(锐角).
∵PB⊥底面ABC,CM?平面ABC
∴CM⊥PB,
∵CM⊥AB,PB∩AB=B,∴CM⊥平面PAB,
∵GM?平面PAB,∴CM⊥GM,
而CM为平面CMG与平面ABC的交线,
又AM?底面ABC,GM?平面CMG,∴∠AMG为二面角G-CM-A的平面角
根据条件可知AM=
,AG=
PA=,
在△PAB中,cos∠GAM=
=,
在△AGM中,由余弦定理求得MG=
,∴cos∠AMG=
,
故平面ABC与平面PEF所成角的二面角(锐角)的余弦值为
.