From 2 ω ω ω = π, ω = 2 is obtained;

ω again? π 12+φ=2kπ+π2, k∈Z, φ∈(0, π2) gives φ=π3,

∴f(x)=2sin(2x+π3)；

(ⅱ)∵f(α2-5π 12)= 2 sin[2(α2-5π 12)+π3]= 2 sin(α-π2)=-2 cosα= 12 13，

∴cosα=-6 13, and from α∈ (π, 3π2), sin α < 0.

∴sinα=- 1？ cos2α=- 1？ (? 6 13)2=- 133 13,

∴f(