(1)f(x)=(ax+b)/(x^2+1)是奇函数,f(-x)=-f(x)(-ax+b)/(x^2+1)=- (ax+b)/(x^2+1),-ax+b=-ax-b, b=-b,所以b=0.又f(1/2)=2/5,所以(a/2)/(1/4+1)=2/5,a=1.∴f(x)=x/(x^2+1). (2)f(0)=0,化为f(t-1)<-f(t)又f(x)是奇函数∴f(t-1)由已知得-1-1<-t<1t-1<-t解得t∈(0,1/2)