(1)f(-x)=loga#[(-mx-1)/(1+x)]=-f(x)=loga#[(1-x)/(mx-1)]=loga#[(x-1)/(1-mx)],故m=-1
(2)f(x)=loga#[(x+1)/(x-1)],(x+1)/(x-1)>0,即x>1或x<-1.在x>1时为减函数,证明如下：
当a>1时,设1=loga#[(x2+1)(x1-1)/(x1+1)(x2-1)],因[(x2+1)(x1-1)/(x1+1)(x2-1)]-1=2(x1-x2)/(x1+1)(x2-1)<0
即(x2+1)(x1-1)/(x1+1)(x2-1)<1,故f(x2)-f(x1)1时f(x)是减函数.