1), which is established by having f( 1+x)=f( 1-x) for any real number x, we can see that:

The symmetry axis of quadratic function f (x) =-x 2+ax+b 2-b+1is: x=a/2= 1, and a=2.

When x ∈ [- 1], f (x) > 0, that is

-x^2+2x+b^2-b+ 1>； 0,

b^2-b+2>； (x- 1)^2。

When x ∈ [- 1, 1], 0

F (x) > 0 holds, then: b 2-b+2 > 4,

Namely B2-b-2 >; 0，(b-2)(b+ 1)>0,

So b <-1, or b >；; 2。

So the value range of b is: b; 2。

2), from the title:-3,2 is the two zeros of the function f (x) = ax 2+(b-8) x-a-ab,

That is, x=-3 and x=2 are quadratic equations: two of ax 2+(b-8) x-a-ab = 0,

So -3+2=-(b-8)/a，-3*2=(-a-ab)/a，

Solution: a=-3, b=5.

The function f (x) =-3x 2-3x+ 18.

( 1)，f(x)=-3x^2-3x+ 18=-3(x+ 1/2)^2+75/4，

The function f(x) monotonically increases at (-infinity,-1/2) and monotonically decreases at (-1/2,+infinity).

And f(0)= 18, f( 1)= 12.

Therefore, the range of f(x) on [0, 1] is: [12,18];

(2), from-3x 2+5x+c < = 0, so:

c & lt=3x^2-5x=3(x-5/6)^2-25/ 12

If the solution set of inequality is r, then: c

So the range of real number c is: c