1
∵点p（1,-根号3）在角a的终边上,
∴tana=-√3
∴f(a)=√3sin2a-2sin²a
=(2√3sinacosa-2sin²a)/(sin²a+cos²a)
=2(√3 tana-tan²a)/(tan²a+1)
=2(-3-3)/(3+1)=-3
2.
f（x）=√3sin2x-2sin²x
=√3sin2x+cos2x-1
=2sin(2x+π/6)-1
∵x∈[-π/6,π/3],
∴(2x+π/6)∈[-π/6,5π/6]
∴sin(2x+π/6)∈[-1/2,1]
2sin(2x+π/6)∈[-1,2]
∴f(x)的值域[-2,1]