21.(1)f(x+y)=f(x)*f(y),令x=y=0得唤好f(0)=[f(0)]^2,∴f(0)=0或1.若f(0)=0,则f(x)=f(x+0)=f(x)*f(0)=0,这谈链慧与“x∈(0,+∞)时含答f(x)∈(1,+∞)”矛盾,∴f(0)=1.(2)f(x)=f(y+x-y)=f(y)*f(x-y),f(y)≠0,∴f(x-y)=f(x)/f(y).