Solution: (1) point A is on the vertical line of line segment OB, then the abscissa of point A is 2, tg∠AOB=2, then the ordinate of point A is 4, and the coordinates of point A are (2,4).

Substituting the coordinates of point A into the hyperbolic equation, we get K=8, and the hyperbolic equation is y = 8/x.

(2) The vertex coordinate of parabola y=(x+m)2+n is (-m, n), the equation of straight line OA is y=2x, and the point m is on OA, then n=-2m. If the straight line PA is perpendicular to the X-axis and intersects the parabola at point P, the coordinate of point P is (2, (m+2)2+n), and PC=(m+2)2+n). When the area of △OBP is S = OB * PC/2 = 4 * (m+2) 2+N)/2 = 2 (m2+2m+4) = 2 (m+1) 2+6, it is obvious that the coordinates of point M are (-1).

(3) Suppose △AMP∽△AMC has AP/AM=AM/AC, that is, AM2=AP*AC①. AM=AO-MO，AO=2，

MO=m (calculated from the coordinates of point m), AM=(2-m). AC=4, AP=AC-PC=4-((m+2)2-2m))=-m2-2m. Substitute AM, AP and AP into ① to get 9m2- 12m+20=0. There is no solution to this equation, so there is no such point.