Solution: (1) f(x) is increasing function's statement f(x) (-2x 2+2ax+4)/(x 2+2) 2 > =0 in the interval [- 1, 1].

That is, -2x 2+2ax+4 >: =0 holds in the interval [- 1, 1].

Then f (- 1) >: =0 f( 1)>=0 means-1

(2) arrange f(x)= 1/x into a quadratic function in the form of x 2-ax-2 = 0, two of which are x 1, x 2.

Then x 1+x2=a x 1x2=-2, so there is

|x 1-x2i^2=(x 1+x2)^2-4x 1x2=a^2+8

So m2+TM+1>; =|x 1-x2| is m 2+TM+1> = √ (a 2+8).

m2+TM+ 1 >； = √ (A 2+8) holds for any A belonging to [- 1, 1] and any T belonging to [- 1, 1].

Then you need the left f (t) of the above formula = mt+m 2+1,and the minimum value must be >; = Maximum value of f(a) on the right.

And f(t) is a linear function, which needs to be discussed.

When m>0, the minimum value f (t) = f (-1) = m 2-m+1> m > = 3; =2

When the minimum value at m<0 is f (t) = f (1) = m 2+m+1> m.

When m=0, it is obviously not true.

So the range of m is m & gt=2 or m.