f(-25)<f(80)<f(11)
解:∵f(x)是奇函数且f(x-4)=-f(x),
∴f(x-4)=f(-x),f(0)
∴f(-25)=f(21)=-f(17)=f(13)=-f(9)=f(5)=-f(1)
f(80)=-f(76)=f(72)=-f(68)=f(64)=-f(60)=f(54)=..=-f(0)
f(11)=-f(7)=f(3)=-f(-1)=f(1)
又∵函数在区间[0,2]上是增函数
0=f(0)<f(1)
∴-f(1)<f(0)<f(1)
∴f(-25)<f(80)<f(11)
故答案为:f(-25)<f(80)<f(11)