∵ In Rt△OAB, tan∠OAB=OBOA= 13.

∴OA= 1，

∴ Point A (- 1, 0)

By substituting the parabola y=-x2+bx+c passing through points a and b, we get:

0=? 1? b+c3=c，

∴b=2,c=3，

The expression of parabola is y=-x2+2x+3.

∫y =-x2+2x+3 =-(x- 1)2+4，

∴ The coordinate of vertex D is (1, 4);

(2) The symmetry axis of parabola is a straight line L = x= 1,

From the meaning of the question: the symmetry point of point B about straight line L is C,

∴ The coordinates of point C are (2,3), and point E (1, 3) is the midpoint of BC.

∴DE= 1，

Point d is the center of gravity of △ △PBC,

∴PD=2DE=2，

That is, PE=3,

Starting from the point p on the straight line l,

∴ The coordinate of point P is (1, 6);

(3)∫P( 1，6)，D( 1，4)，

∴PD=2, you can see that the parabola y=-x2+2x+3 moves up by 2 units.

The expression after parabola translation is y=-x2+2x+5.

The coordinate of point m is (m, n).

The heights of PD on both sides of △MPD and △BPD are |m- 1| and 1 respectively.

Therefore, the area of △MPD is equal to twice the area of △BPD,

| m- 1 | = 2。

Solution: m 1=- 1, m2 = 3.

The point m is on the parabola y=-x2+2x+5,

∴n 1=2,n2=2，

∴ The coordinates of point m are M 1 (- 1, 2) and M2 (3, 2) respectively.