1、证明：
|a|=1,|b|=1,
∴(a+b)*(a-b)
=a^2-b^2
=1-1
=0
∴(a+b)⊥(a-b)
2、|a-b|^2
=a^2+b^2-2ab
=1+1-2cos2x
=4-4(cosx)^2
=1/9
∴(cosx)^2=35/36,
又∵x∈[-π/3,π/2]
∴cosx=√35/6
3、f(x)
=ab+2|a+b|
=cos2x+2+2cos2x
=2+3cos2x
∵x∈[-π/3,π/2]
∴cos2x∈[-1,1]
∴f(x)∈[-1,5]
∴f(x)的最小值是-1.