Solution: (1). OABC is a square because there is no chart. According to the meaning of the question, the coordinates of the four vertices are 0 (0,0); A(0，2)；

B(2，2)； C(2，0)； Parabola y=ax? +bx+c passes through a, b, e (-2/3,0), so there is an equation:

c=2.............................( 1)

4a+2b+c=2.................(2)

(4/9)a-(2/3)b+c=0.......(3)

A=-9/8, b=9/4 and c = 2； are obtained by simultaneous solution of three formulas.

Therefore, the quadratic resolution function is y=-(9/8)x? +(9/4)x+2

(2)。 The equation of the straight line BE is y = [2/(2+2/3)] (x+2/3) = (3/4) (x+2/3), that is, 3x-4y+2 = 0............(4);

The equation of circle D with diameter OC is (x- 1)? +y？ = 1; Because the distance D (1, 0) be d = 5/√ (9+16) = 5/5 =1= circle.

Radius d, so BE is the tangent of circle d.

(3)。 Parabola y=-(9/8)x? The symmetry axis of +(9/4)x+2 is x =1; Substituting into equation (4) gives y=5/4, that is, the coordinate of point P is (1, 5/4);

Let the coordinates of m be (2, t), (0

That is 3x-4y-6+4t=0..........(5).

Let y=0 in formula (5), that is, x=2-(4/3)t, then the coordinate of point n is (2-(4/3)t, 0);

∣MN∣=√[( 16/9)t？ +t？ ]=(5/3)t；

The distance from point P to a straight line (5), that is, the height of △PMN on the side Mn = ∣ 3-5-6+4t ∣/5 = ∣ 4t-8 ∣/5 = (8-4t)/5;

So the area of △PMN is s = (1/2) × [(5/3) t ]× [(8-4t)/5] = (1/3) (4t-2t? )、(0 & ltt & lt2)。

S= -(2/3)t？ +(4/3)t= -(2/3)(t？ -2t)=-(2/3)[(t- 1)？ - 1]= -(2/3)(t- 1)？ +2/3≦2/3;

That is, when t= 1, s gets the maximum value of 2/3.