1) bring x=- 1 into the equation to get f(- 1)= 1, and get a= 1. The original function can be changed to: f (x) = x 3-3x- 1.

The derivative function f' (x) = 3 (x+ 1) (x- 1), which shows that the function increases in (-∞,-1), [1, ∞) and increases in [-/kloc-0].

2) The problem is transformed into the tangent problem of the function, that is, the position of the straight line is guaranteed to be between two tangents with the original function slope of 9.

The derivative function f' (x) = 3 (x+ 1) (x- 1), so that f' (x) = 9, x = 2, f (-2) =-3, f (2) =1;

Therefore, the two tangent equations of the function are y=9x+ 15 and y=9x- 17. So the range of m is (-17, 15).