The point p(3,-1) is on f (x) = m+logax relative to the symmetrical point Q (1,-1) of the straight line X = 2.

∴- 1 = m+log 1, which means m=- 1.

Solvable a=2.

( 1)f(x)=- 1+log2x，g(x)+ 1 =- 1+log2(x+ 1)

∴g(x)=-2+log2(x+ 1)

(2)h(x)=g(x*2)-f(x)=-2+log2(x^2+ 1)+ 1-log2x=- 1+log2[(x^2+ 1)/x)

That is, h (x) =-1+log2 (x+1/x).

∵log2x monotonically increases, ∴ Only when x+ 1/x takes the minimum value, the function value is the minimum value.

For the "tick" function, the minimum value is taken when x= 1/x, and the minimum value is 2 at this time.

H(x)min=- 1+log22=0，x= 1。

2.f'(x)= 1/x-2x+ 1

Let f'(x)=0 to get -(2x 2-x- 1)/x = 0.

∴x=- 1/2 or x= 1, obviously the domain specifies x >；; 0, so there is only one zero.

-(2x^2-x- 1)/x>； 0 to get x

(1) monotonically decreasing interval [1, +∞), monotonically increasing interval (0, 1)

(2) If

If a> 1, the maximum value is obtained at x= 1. f(x)max=f( 1)=2