C= 1 from f(0)= 1。

And f(x+ 1)-f(x)=2x.

2ax+a+b=2x

∴a= 1 b=- 1

∴f(x)=x^2-x+ 1

(2)∫ inequality f (x) > 2x+m holds for any x belonging to-1.

The inequality f (x)-2x > m holds for any x that belongs to-1.

∴m<； [f(x)-2x] minimum value

∵[f(x)-2x]min=[x^2-3x+ 1]min=- 1

∴m<； - 1

2.( 1) solution: let x=y=0, then there is f(0+0)=f(0)+f(0).

∴f(0)=0

(2) Prove that if x=-y, then f(x-x)=f(x)+f(-x).

∴ f(-x)=-f(x)

(3) Solution: ∫f( 1)= 1

∴f(2)=f( 1+ 1)=f( 1)+f( 1)=2

∴f(a- 1)+2=f(a- 1)+f(2)=f(a+ 1)

∫ f (2a) > f (a-1)+2 = f (a+1), and the function f(x) is a decreasing function on R.

∴2a>； a+ 1

∴a>； 1