（1）、∵f(x)是奇函數(shù)
∴f(-x)=log2^[(-x-1)/(1+ax）]=-log2^[(x-1)/(1-ax)]
∴（-x-1)/(1+ax)=[(x-1）/(1-ax)^(-1)
化簡(jiǎn)得：(a^2-1)x^2=0
a=±1
（2）、設(shè)x2>x1>1
則f(x2)-f(x1)=log2^[(x2-1)/(1-ax2)]-log2^[(x1-1)/(1-ax1)=log2^{[(x2-1)/(1-ax2)]/[(x1-1)/(1-ax1)}
log2^[(x2-1)(ax1-1）]/[ax2-1)(x1-1)]
∵x2>x1>1,∴x2-1>0,x1-1>0；ax2-1>1,ax1-1>1；(x2-1)/(x1-1)>1
∴l(xiāng)og2^[(x2-1)(ax1-1）]/[(x1-1)(ax2-1）]>0；即：f(x2)>f(x1)
∴f(x)在x∈（1,+∞）上單調(diào)遞增.