Substituting x=0 into y=(x-2)2-2, we get: y=2,

The coordinates of point B are (0,2).

(2) Extending EA, intersecting with Y axis at point F,

AD = AC，∠AFC=∠AED=90，∠CAF=∠DAE，

∴△AFC≌△AED，

∴AF=AE，

∫ point A(m, -m2+m), point B(0, m),

∴AF=AE=|m|,BF=m-(-m2+m)=m2，

∠∠ABF = 90-∠BAF =∠DAE，∠AFB=∠DEA=90，

∴△ABF∽△DAE，

∴ bfaf = aede, that is, m2 | m | = | m | de.

∴DE= 1.

(3) ① The coordinate of point A is (m, -m2+m),

∴ The coordinate of point D is (2m, -m2+m+ 1),

∴x=2m,y=-m2+m+ 1，

∴y=-(x2)2+x2+ 1，

The analytical formula of the ∴ function is: y =-14x2+12x+1

② If PQ⊥DE is at point Q, then △ DPQ △ BAF,

(i) When the quadrilateral ABDP is a parallelogram (as shown in figure 1),

The abscissa of point p is 3m,

The ordinate of point p is: (-m2+m+1)-m2 =-2m2+m+1,

Substituting the coordinates of p (3m, -2m2+m+ 1) into y =-14x2+12x+1gives:

-2 m2+m+ 1 =- 14×(3m)2+ 12×(3m)+ 1，

Solution: m=0 (A, B, D and P are on the same line at this time, so give up) or M = 2.

(ii) When the quadrilateral ABPD is a parallelogram (as shown in Figure 2),

The abscissa of point p is m,

The ordinate of point p is: (-m2+m+ 1)+m2=m+ 1,

Substituting the coordinates of P(m, m+ 1) into y =-14x2+12x+1gives:

m+ 1 =- 14m 2+ 12m+ 1，

Solution: m=0 (A, B, D and P are in the same line at this time, so we should abandon them) or m=-2.

To sum up: the value of m is 2 or -2.