非奇非偶:{0,1,2,3,4,5,9}
解:(1)∵f(-1.6)=[(-1.6)?[-1.6]]=[3.2]=3,f(1.6)=[1.6?[].6]]=[1.6]=1≠f(-1.6),
f(1.6)≠-f(-1.6),
∴函数f(x)=[x[x]](x∈R)为非奇非偶函数.
(2)x=-2,f(-2)=[-2?[-2]]=4,-2<x<-1.5,f(x)=[x[x]]=3,,当-1.5≤x<-1,f(x)=[x[x]]=2,同理可得x=-1,f(-1)=[-1?[-1]]=1,-1<x<1,f(x)=[x[x]]=0,1≤x<2,f(x)=[x[x]]=1,2≤x<2.5,f(x)=[x[x]]=4,2.5≤x<3,f(x)=[x[x]]=5,f(3)=9.
故答案为:非奇非偶;{0,1,2,3,4,5,9}.