The domain of F(x)=f(x)+g(x) is x > 0.

Because a < 0

So f' (x) = 1/x-a/x? >0

So F(x) is increasing function at (0, +∞).

2) F'(x)= 1/x-a/x？ =-a( 1/x？ - 1/ax+ 1/4a？ )+ 1/4a =-a( 1/x- 1/2a)？ + 1/4a

So the function f' (x) of the symmetry axis x=2a is increasing function at (0,3).

f(3)= 1/3-a/9≤ 1/2a ≥- 3/2。

So the range of a is -3/2.0)

3) let t=x? +1, so t≥ 1

y=g(2a/(x^2+ 1)+m- 1=(x？ + 1)/2+m- 1 = t/2+m- 1

y=f( 1+x^2)=ln(x？ + 1)=lnt

Let P(t)=t/2+m- 1-lnt.

P'(t)= 1/2- 1/t

So when t∈(2, +∞) is a increasing function t∈ irreducible function.

Because the image of y = g (2a/(x 2+1)+m-1) and the image of y = f (1+x 2) have exactly four different intersections, there are two different roots when P(t)=0.

So there is p (1) ≥ 0 p (2) < 0.

M > 1/2+M- 1-ln 1 > 0。

1+m- 1-LN2 < 0 m < LN2。

So the range of m is (1/2,2).