(1) Analysis: According to the meaning of the question, the quadrangular cone P-ABCD has a right trapezoid BC//AD, AB⊥AD, PA⊥ base.

Connect BD, AC, pay F；; FE//PC exceeds F, and E exceeds PA.

∵EF∈ surface EBD, ∴PC// surface EBD, then E is what you want.

PA = AD = 2，AB=BC= 1

∴AC=√2,BD=√5

⊿adf∽⊿cbf==>； AF/FC=AD/BC=2

∴af/f=ae/ep=2==>； (AF+FC)/FC=(AE+EP)/EP=3

∴PE/PA= 1/3

(2) Analysis: Establish a spatial rectangular coordinate system A-xyz with A as the origin, AD as the X axis, AB as the Y axis and AP as the positive direction of the Z axis.

Then point coordinates:

A(0，0，0)，B(0， 1，0)，C( 1， 1，0)，D(2，0，0)，P(0，0，2)

Vector PB=(0, 1, -2)

Vector PC=( 1, 1, -2)

Vector PD = (2,0, -2)

Let the vector m=(x, y, z) be the normal vector of surface PBC,

Vector m vector PB=y-2z=0

Vector m vector PC=x+y-2z=0

Let y= 1, then x = 0 and z = 1.

Vector m = (0,2,1) = = > Vector m|=√5

Let the vector n=(x, y, z) be the normal vector of surface PCD.

Vector n Vector PD=2x-2z=0

Vector n vector PC=x+y-2z=0

Let x= 1, then y = 1 and z = 1.

Vector n = (1, 1,1) = >; Vector n|=√3

Vector m vector n=3

Cos< vector m, vector n > = vector m vector n/| vector M || = 3/√15 = √15.

The cosine of dihedral angle B-PC-D is -√ 15/5.