Let OA=m, then OB=OC=5m, AB=6m,

From △ABC= AB×OC= 15, we get ×6m×5m= 15, and the solution is m= 1 (excluding negative values).

∴A(- 1,0),B(5,0),C(0,-5)，

Let the analytical formula of parabola be y=a(x+ 1)(x-5), and substitute it into the coordinates of point C to get a= 1.

The analytical formula of parabola is y=(x+ 1)(x-5),

That is y = x2-4x-5;

(2) Let the coordinate of point E be (m, m2-4m-5) and the parabola axis of symmetry be x=2.

From 2(m-2)=EH, 2(m-2)=-(m2-4m-5) or 2(m-2)=m2-4m-5,

The solution is m = 1 or m = 3.

∵ m > 2, ∴m= 1+ or m=3+,

Side length EF=2(m-2)=2 -2 or 2+2;

(3) existence.

According to (1), OB=OC=5,

∴△OBC is an isosceles right triangle, and the analytical formula of straight line BC is y=x-5.

According to the meaning of the question, the distance between the straight line y=x+9 or the straight line y=x- 19 and BC is 7.

Combining linear analytical formula and parabolic analytical formula, the coordinates of m point are obtained.

The coordinates of point ∴M are (-2,7), (7, 16).