f(2)=8+4a+2b=2，①

f'(2)= 12+4a+b= -3，②

A =-6 and b = 9 can be obtained from the above two formulas.

From f' (x) = 3 (x- 1) (x-3), it is easy to know that the function increases at (-∞, 1), decreases at (1, 3) and increases at (3, +∞).

(2) f(x)=x(x-3)？ ,

For any real number x, there are

f(2+x)+f(2-x)

=(2+x)(x- 1)？ +(2-x)(x+ 1)？

= 4, so the function image is symmetrical about point (2, 2).

(3) Let m = ∑ (I = 1, 8067) f (I/20 17),

Then m = ∑ (I = 1, 8067) f [(8068-I)/20 17], (in reverse order)

So 2m = ∑ (I = 1, 8067) 4.

=4×8067=32 188,

So the original formula = 16094.