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f(x)=2sin(11π/6 -2x)+3+m=2sin(-π/6 -2x)+3+m=-2sin(2x +π/6 )+3+m

(1)因为-sinx的递增区间为2kπ+π/2≤x≤2kπ+3π/2,k∈Z,

所以2kπ+π/2≤2x +π/6≤2kπ+3π/2,k∈Z,

解出x得单增区间（亲,请写成集合或者区间形式）.

(2) π/2≤x≤π,

7π/6≤2x +π/6≤13π/6,

-1≤sin(2x +π/6)≤1/2

2+m≤f(x)≤5+m,

2+m=2, and 5+m=5,

m=0.