∴-ωπ/2≦ωx≦π/3 is increasing function, and the monotonic increasing interval of f(x)=sinx is [2kπ-π/2, 2kπ+π/2].

∴-ππ/2≧2kπ-π/2 and 2ππ/3≦2kπ+π/2, the solution is ω≦3/4, and ω >; 0,∴0<； ω≦3/4 。

(2)f(x)= 2 sinx+ 1 and f (x) = cosx+ 1, ∴ 2sinx+ 1 = cosx+ 1, that is, tanx = 1.

tan2x=2tanx/( 1-tan？ x)= 1/( 1- 1/4)= 4/3，

∴tan(2x+π/6)=(tan2x+√3/3)/( 1-√3/3tan2x)=(4/3+√3/3)/( 1-√3/3×4/3)=(48+25√3)/ 1 1