Solution: ∫f(x)= 2x 3/3+ax 2+ 1? ∴f'(x)=2x^2+2ax=x(2x+2a)？ Let f'(x)=0? Then x=0 or x=-a, ∫ has two extreme points, ∴a≠0.

( 1)

At this time f (x 1) = 1, and f (x2) =-2a3/3+a3-a2+1= a3/3-a2+1.

f(x 1)+f(x2)=a^3/3-a^2+2

(2) when a >; 0，x 1=-a，x2=0，f (x 1)+f (x2) = a 3/3-a 2+2。

Let G (a) = a 3/3-a 2+2, then G' (a) = a 2-2a = a (a-2), and let g'(a)=0? Then a=0 or a=2, when a; At 2 o'clock, g'(a)>0.g(a) increases monotonically, ∴a=0 is the maximum value of g(a), g(0)=2, a=2 is the minimum value of g(a), and g(2)=2/3, ∫ the minimum value is 2. If a-2 is extracted, a3-3a2+4 = (a-2) (a2-a-2) = (a-2) 2 (a+1) = 0, so the other root is a=- 1. 2/3, then the value range of a is a >；; -1 and a≠2.