∴AC⊥BD in O, O as OZ⊥ bottom ABCD.

Establish a spatial rectangular coordinate system O-xyz with O as the origin, OA as the X axis, OD as the Y axis and OZ as the positive direction of the Z axis.

Let AB=2

Point coordinates: O (0,0,0), A C(-√3 3,0,0), B (0 0,0 1 0), C (-√ 3,0,0), D (0,0 1 0), A65438.

(1) vector ed1= (0,2,2-z), vector ca = (2 √ 3,0,0), vector D 1A=(√3,-1, -2).

∵D 1E⊥ surface D 1AC, ∴ vector ed1* ca = 0; * ca = 0； Vector ED 1*D 1A=0== >-2-2(2-z)=0== >z = 3；

∴E(0,- 1,3)

∵d 1a=d 1c,∴d 1o⊥ac,d 1o=√(do^2+dd 1^2)=√5

∵ea=ec,∴eo⊥ac,eo=√(bo^2+be^2)=√ 10

Vector ED 1|=√5

According to the cosine theorem cos ∠ EOD1= (OE 2+OD12-ED12)/(2oe * OD1) = √ 2/2.

∴ dihedral angle D 1-AC-E is 45;

(2) suppose there is a little p on D 1E, so that A 1P// faces EAC.

Let D 1P/PE=λ.

∴ vector D 1P=λ vector PE=λ vector (D 1E-D 1P)

We can know P (0, (1-λ)/( 1+λ), (2+3 λ)/( 1+λ)) from the coordinates of the fixed point.

∴ vector d 1p = (0, -2λ/( 1+λ), λ/( 1+λ))

∴ vector A 1p = (-√ 3, (1-λ)/( 1+λ), λ/( 1+λ))

The vector EA =(√3 1, 3) and the vector EC =(√3 1, 3).

Let the vector m=(x, y, z) be the normal vector of face EAC.

Vector EA*m=√3x+y-3z=0.

Vector EC*m=-√3x+y-3z=0.

Let z= 1 and get the vector m = (0,3, 1).

∫a 1p∑ plane EAC, ∴ vector A 1P⊥ vector m.

∴ vector A 1P* vector m = 0+3 (1-λ)/(1+λ)+λ/(1+λ) = 0.

∴λ=3/2

∴ there is a little p, which makes a 1p∑ plane EAC, D 1P/PE=3/2.