According to the image, from -π/3 to 2π/3 is a half period, so t = [(2π/3-(-π/3)] * 2 = 2π.

That is 2π/ω=2π, so ω= 1.

So f(x)=2sin(x+φ)

Substitute the highest point (2π/3,2) (or the lowest point (-π/3,2)) into the function to get 2=2sin(2π/3+φ).

So sin(2π/3+φ)= 1.

So 2π/3+φ = π/2+2kπ (k ∈ z),

That is φ = 2kπ-π/6 (k ∈ z)

Because -π/2

So φ =-π/6

So f (x) = 2 sin (x-π/6)

(2) Because f(a)=3/2, that is, sin (a-π/6) = 3/4.

So sin(2a+π/6)=cos[π/2 -(2a+π/6)] (the inductive formula cos(π/2-a)=sina is used here).

=cos(π/3-2a)=cos(2a-π/3) (the inductive formula cos(-a)=cosa is used here).

= cos [2 (a-π/6)] =1-2 [sin (a-π/6)] 2 (here, the formula of twice the angle is used).

= 1-2(3/4)^2=- 1/8

That is sin(2a+π/6)=- 1/8.