f(x)=-2sin(x-π/3);
1)最小正周期T=2π;
2)由2kπ-π/2≦x-π/3≦2kπ+π/2;
得2kπ-π/6≦x≦2kπ+5π/6;为单调减区间
3)当x-π/3=2kπ-π/2,即x=2kπ-π/6时f(x)取得最大值,为2