(1)
f(x) = log(3){ (a+x)/(a-x) }
(-1,1)
f(-1) = log(3){ (a-1)/(a+1) } =1
(a-1)/(a+1) = 3
a-1 = 3a+3
a=-2
(2)
f(x/2)+f(2/x)= 2f(1)
log(3){ (-2+x/2)/(-2-x/2) } + log(3){ (-2+2/x)/(-2-2/x) } = 2log(3){ (-2+1)/(-2-1) }
log(3){[(-4+x)/2][2/(-4-x)].[(-2x+2)/x][x/(-2x-2)] = 2log(3)(1/3)
(-4+x)(-2x+2)/[(4+x)(2x+2)] = (1/3)^2
9(-4+x)(-2x+2) = (4+x)(2x+2)
-72+90x-18x^2=8+10x+2x^2
20x^2-80x+80=0
x^2-4x+4 =0
(x-2)^2 =0
x =2