Substituting the coordinates of a, b and c, we can get: 16a+4b+c = 04a? 2b+c=0c=4，
Solution: a =? 12b= 1c=4。
So the analytical formula of parabola is: y =- 12x2+x+4.
(2) When passing through point M, MC⊥OA at point C',
The coordinate of point M is (x,-12x2+x+4).
Then s quadrilateral BOAM=S trapezoid BOC ′ m+s △ MC ′ a =12 (bo+c ′ m) × oc ′+12ac ′× c ′ m =12 (4-12x2+)
S△AOB= 12OB×OA=8，
So S△AMB=S quadrilateral BOAM-S△AOB=-x2+4x=-(x-2)2+4.
Therefore, when x=2, that is, the coordinate of point M is (2,4), the area of △AMB is the largest, and the maximum value is 4.
(3)
Make a straight line y=-x, if ∠ 0 = 90 is in a right-angled trapezoid with OB as the base, then point P coincides with point C,
Then the coordinate of point Q at this time is (-2,2);
If ∠ b = 90 is in a right-angled trapezoid with the base of OB,
If point b is perpendicular to OB, the intersection of parabolas is the position of point p,
The q coordinate of this point is (2, -2).