(2)① According to △AOM∽△PED, get DE: PE: PD = 3: 4: 5, then get PD=yP-yD, and get the maximum of two functions;

② When the G-point falls on the Y-axis, from PC=AO=2 of △ ACP△ GOA, that is,-14x2-34x+52=2, the solution is X =-3 172.

So P1(-3+172,2) and P2 (-3-172,2), when point F falls on the Y axis, P3(-7+892, -7+892) and P4 (can be obtained in the same way.

∴ The coordinates of point A are (2,0), and the coordinates of point B are (-8,-152).

Parabola y=- 14x2+bx+c passes through point a and point b,

de { 0 =- 1+2 b+ C- 152 =- 16-8 b+ C。

The solution is b=-34 and c = 52.

∴y=- 14x2-34x+52.

(2)① Let a straight line y=34x-32 intersect with the Y axis at point m,

When x=0, y =-32. ∴ om = 32。

∫ The coordinate of point A is (2,0), ∴ OA = 2. ∴ AM = OA2+OM2 = 52。

∵OM:OA:AM=3:4:5。

Judging from the meaning of the question, ∠PDE=∠OMA, ∠ AOM = ∠ PED = 90, ∴△AOM∽△PED. ..

∴DE:PE:PD=3:4:5.

Point p is a moving point on the parabola above the straight line AB,

∵PD⊥x axis,

∴PD two points have the same abscissa,

∴pd=yp-yd=- 14x2-34x+52-(34x-32)

=- 14x2-32x+4

∴l= 125(- 14x2-32x+4)

=-35x2- 185x+485。

∴l=-35(x+3)2+ 15.

When ∴x=-3, the maximum value of L = 15.

② There are three points P that satisfy the meaning of the question, namely P1(-3+172,2) and P2 (-3-172,2).

P3(-7+892，-7+892)。 Comments: This question mainly examines the second letter.