A<x<b for f (x) = f (3x-1)-f (3x+1), then

a & lt3x- 1 & lt； b a & lt3x+ 1 & lt； simplify

(a+ 1)/3 & lt； x & lt(b+ 1)/3(a- 1)/3 & lt； x & lt(b- 1)/3

By B-A > 2 is also b-1>; A+ 1 means b+1>; b- 1 & gt； a+ 1 & gt； a- 1

(a+ 1)/3 & lt； x & lt(b- 1)/3

(2)f(x 1+X2)= f(x 1)+f(X2)，f(8)=3

Then f(8)=f(4+4)=f(4)+f(4)=2*f(4)=3, then f(4)=3/2.

F(4)=f(2+2)=2*f(2)=3/2，f(2)=3/4。

(4) If f (x+2) = 1/f (x) and f( 1)=-5, then f(5)=f(3+2)= 1/f(3).

And f (3) = f (1+2) =1/f (1) =-1/5.

f(5)=-5

(5)g(x)= 1-2x f[g(x)]= 1/x^2-x^2

If f( 1/2) is needed, as long as g(x)= 1/2 = 1-2x, then substitute x= 1/4 into 1/x 2.

f( 1/2)= 16- 1/ 16 = 255/ 16