(1) 令2kπ-π/2≤x+π/3≤2kπ+π/2,解得：2kπ-5/6*π≤x≤2kπ+π/6
所以单调递增区间为：[2kπ-5/6*π,2kπ+π/6] (k∈Z)
(2)当π/6≤x≤π/2时,π/2≤x+π/3≤5/6*π,1/2≤sin(x+π/3)≤1,即值域为：[1/2,1]