如图,正四棱柱ABCD-A1B1C1D1中,AA1=2AB=4,点E、F分别为BC,CD的中点,点G在棱CC1上（Ⅰ）是

1、要A1G⊥平面EFG,则A1G⊥EG,A1G⊥FG,根据勾股定理,A1G^2+EG^2=A1E^2,
A1G^2+FG^2=A1F^2,
设CG=x,
AE=√5,A1E^2=AE^2+AA1^2=AB^2+BE^2+AA1^2=21,
A1G^2=A1C1^2+C1G^2=8+(4-x)^2,
EG^2=CG^2+CE^2=x^2+1
8+(4-x)^2+x ^2+1=21,
x^2-4x+2=0,
∴x=2±√2,
即G有两点,CG=2±√2.
2、若二面角A1-FG-E的余弦值为零,则两平面相垂直,则A1G⊥平面EFG,
由上所述CG=2±√2,
S△EFC=S△CBD/4=（2*2/2）/4=1/2,
∴VG-CEF=S△EFC*CG/3=（1/2）*（2±√2）/3=（2±√2）/6.