From f(a+b)=f(a)f(b),

f(x+ 1)=f(x)f( 1)，

and

When x>0, f(x)> 1, so f( 1)> 1, so

F(x+ 1)>F(x), which proves that f (x) is a increasing function on R.

2.

Since any A and B belong to R, there is f(ab)=af(b)+bf(a).

F (x) = f (1× x) = f (x)+xf (1), so f( 1)=0.

f(-x)= f((- 1)×x)=-f(x)+xf(- 1)

and

F (1) = f (-1) (-1)) =-2f (-1) = 0, and it is deduced that f (-1) = 0;

So f(-x)=-f(x), and f(x) is odd function.

3.

f(x)=|2x-4|+ 1

Write it as a piecewise function,

When x≥2, 2x-4≥0 and f(x)=2x-3.

When x < 2 and 2x-4 < 0, f(x)=5-2x.

F(x)≤ax has a solution.

When x≥2 and x≤3/(2-a), there is a solution of 3/(2-a)≥2 in this part; when a≥0.5 and a=0.5, there is only one solution of x=2.

When x

To sum up, the value range of a is a≥0.5.

4.

First, find all the functions and analyze their properties:

F(x) is odd function f(-x)=-f(x)=-2x-x? =2(-x)-(-x)？

When x > 0, f(x)=2x-x? And x=0, f(x)=f(-x)=0, and the function is uninterrupted.

It can be seen that when x < 0, f(x)=(x+ 1)? -1 is a parabola with an upward opening, the symmetry axis is x=- 1, and the minimum value at x=- 1 is-1, which is determined by the properties of parabola.

When x

When-1 ≤ x < 0, f(x) is increasing function, and the value range is [- 1, 0].

Similarly, when x > 0, f(x)=-(x+ 1)? +1 is a parabola with a downward opening, the symmetry axis is x= 1, and the maximum value at x= 1 is 1. Therefore,

When x > 1, f(x) is a decreasing function;

When 0 < x ≤ 1, f(x) is increasing function, and the value range is (0, 1).

To sum up, when x is between [- 1, 1], f(x) is increasing function, and its value range is [- 1, 1].

When the domain required in the question is (a, b), the range is (1/b, 1/a).

When a < b ≤- 1 and f(x) is a decreasing function, then f(b)= 1/b and f(a)= 1/a are brought into the solution of the equation.

x？ +2x= 1/x

x？ +2x？ - 1=0

(x+ 1)(x？ +x- 1)=0

Three solutions x =- 1, x = (- 1-√ 5)/2, x = (- 1+√ 5)/2,

a