Answer: Solution: (1) According to the meaning of the question,1+f (1)-3 = 0 ∴ f (1) = 2.

That is 2=

1

three

-a+a2- 1+b，a2-a+b-

eight

three

=0,

∫ The slope of tangent x+y-3=0 is-1,

∴f'( 1)=- 1, namely a2-2a+ 1=0, a= 1,

Substitute into the solution to get b=

eight

three

(ii) Because the function f(x) is not monotonic in the interval (-1, 1),

So the equation f'(x)=0 has a solution at (-1, 1).

Because f' (x) = x2-2ax+a2-1= [x-(a-1)]? [x-(a+ 1)]，

So-1< a-1<1or-1< a+1<1... (10) ∴ a ∈.