(1)解析：∵向量a=(cos(3x/2),sin(3x/2)),b=(cos(x/2),-sin(x/2))
∵a⊥b
∴向量a*b=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)
=cos(2x)=0
==>x∈{x|x=kπ+π/4,k∈Z}
(2)解析：∵向量c=(√3,-1)
向量a-c=(cos(3x/2)-√3,sin(3x/2)+1)
|向量a-c|^2=(cos(3x/2)- √3)^2+(sin(3x/2)+1)^2
=cos^2(3x/2)+3-2√3cos(3x/2)+sin^2(3x/2)+1+2sin(3x/2)
=4sin(3x/2-π/3)+5
|向量a-c|^2最大值为9
∴|向量a-c|最大值为3