∵ quadrilateral ADEF is a square,

∴FA//ED

∴∠CED is the angle formed by non-planar straight lines CE and AF.

∵FA⊥ plane ABCD,

∴FA⊥CD. So ED⊥CD.

In Rt△CDE, CD= 1, ED=2, and CE= 2 (CD 2+ED 2) = 3.

So COS∠CED=ED/CE=2 root number 2/3.

∴ The cosine of the angle formed by straight lines CE and AF on different planes is 2/3.

(2)

(ii) It is proved that the intersection point B is BG//CD and the intersection point AD is G,

Then BGA = CDA = 45

If ∠ bad = 45, you can get BG⊥AB.

So CD⊥AB.

And CD⊥FA,FA∩AB=A,

So CD⊥ aircraft base.

(3)

Solution: From (2) and known, we can get AG= radical number 2,

That is, g is the midpoint of AD.

Take the midpoint n of EF, connect GN, and then GN⊥EF.

BC//AD

∴BC//EF.

If the intersection n is NM⊥EF and BC is at m, then ∠GNM is the plane angle of dihedral angle b-ef-a.

Connect GM, and you can get the AD⊥ aircraft GNM.

Therefore, AD⊥GM. Therefore BC⊥GM.

GM= root number 2/2 is known.

From NG//FA,FA⊥GM, NG⊥GM is obtained.

In Rt△NGM, tan∠GNM=GM/NG= 1/4.

The tangent of dihedral angle B-EF-A is 1/4.