g(x)=-4cos^2(x+π/6)+4sin(x+π/6)-a

1.f(x)=g(x+π/3)+1
=-4cos^2(x+π/3+π/6)+4sin(x+π/3+π/6)-a+1
=-4cos^2(x+π/2)+4sin(x+π/2)-a+1
=-4sin^2(x)+4cosx-a+1
=4-4sin^2(x)+4cosx-a-3
=4cos^2(x)+4cosx-a-3
2.f(x)+8+a=4cos^2(x)+4cosx-a-3+8+a
=4cos^2(x)+4cosx+1+4
=(2cosx+1)^2+4
∵cosx∈[-1,1]
∴2cosx+1∈[-1,3]
∴(2cosx+1)^2+4∈[4,13]
∴y=log(1/2)[f(x)+8+a]
=-log(2)[(2cosx+1)^2+4]
∴y∈[-log(2)13,-log(2)4]
y∈[-log(2)13,-2]
3.f(x)=4cos^2(x)+4cosx-a-3
=4cos^2(x)+4cosx+1-a-4
=(2cosx+1)^2-a-4
当x∈[-π/4,2/3π]时,
cosx∈[-1/2,√2/2]
∴2cosx+1∈[0,√2+1]
∴(2cosx+1)^2-a-4∈[-4-a,2√2-1-a]
∵当x∈[-π/4,2/3π]时,f（x）=0恒有解
∴-4-a≤0≤2√2-1-a
解得：-4≤a≤2√2-1
∴a的取值范围是[-4,2√2-1]