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3.( 12 points) Solution 1: (1) Connect BD, take DC midpoint G, and connect BG.

So DG = GC = BG = 1, that is, △DBC is a right triangle, so BC⊥BD.

And SD⊥ plane ABCD, so BC⊥SD,

So BC⊥ aircraft BDS, BC⊥DE.

For BK⊥EC, k is vertical feet. Because the plane EDC⊥, the plane SBC, the plane EDC, BK⊥DE.DE are perpendicular to the two intersecting lines BK and BC in the plane SBC.

De⊥sb. de⊥ecde⊥ Aircraft Company

SB= =，

DE= =，

EB= =，SE=SB-EB=，

So se = 2eb.

(2) from SA = =, AB = 1, SE = 2eb, AB⊥SA, known AE = = 1, AD = 1

So △ADE is an isosceles triangle.

Take the midpoint f of ED, connect AF, and then AF⊥DE, af = =.

If FG is connected, then fg∨EC, FG⊥DE.

So ∠AFG is the plane angle of dihedral angle A-DE-C. 。

Add AG, ag =, fg = =,

cos∠AFG= =-。

So the dihedral angle A-DE-C is 120.

Solution 2: Take D as the coordinate origin and ray DA as the positive semi-axis of the X axis, and establish the rectangular coordinate system Dxyz as shown in the figure.

Let A( 1, 0,0), then B( 1, 1, 0), C (0,2,0), S (0 0,0,2).

( 1) =(0,2,-2),

=(- 1, 1,0).

Let the normal vector of plane SBC be

n=(a，b，c)，

Take n from n⊥n⊥? =0，n？ =0.

So 2b-2c = 0, -a+b = 0.

Let A = 1, then B = 1, C = 1, n = (1, 1).

Let = λ (λ > 0),

Then e (,,).

=( , ,), =(0,2,0).

Let the normal vector m = (x, y, z) of the plane CDE,

From m⊥, m⊥, get

m？ =0，m？ =0.

Therefore ++= 0, 2y = 0.

Let x = 2, then m = (2 2,0,-λ).

M⊥n，m？ n=0，2-λ=0，λ=2。

So se = 2eb.

(2) We know E (,,) from (1), and if the midpoint of DE is F, then F (,) = (,-,-),

So what? = 0, from which FA⊥DE is obtained.

= (-,-) Here we go again, so? = 0, thus obtaining EC⊥DE,

The angle between vector and is equal to the plane angle of dihedral angle ADEC.

So cos = =-,

So the dihedral angle A-DE-C is 120.