（1）
f(x)=√3/2×sin2x-cos²x-1/2
=√3/2×sin2x-(2cos²x-1)/2-1
=√3/2×sin2x-cos2x/2-1
=sin2x×cosπ/6-cos2x×sinπ/6 -1
=sin(2x-π/6) -1
-1≤sin(2x-π/6)≤1,最小正周期＝π
所以,f(x)的最小值＝-1-1＝-2,最小正周期＝π
（2）
f(C)=0
f(C)=sin(2C-π/6)-1=0
sin(2C-π/6)=1
2C-π/6=π/2
2C=2π/3
C=π/3
sinB=2sinA
所以,b=2a
cosC=(a²+b²-c²)/2ab
所以,1/2=(5a²-3)/4a²
4a²=10a²-6
a²=1
a=1或a=-1(舍去)
b=2a=2
所以,a＝1,b＝2