1.:( 1) f (x) = vector a* vector b = 2 (cosx) 2-1+√ 3sin2x+m+1= 2cos (2x-π/3)+m+6544.

(For a square)

Therefore, the minimum positive period of f(x) is π. If f(x) is monotonically increasing, then 2kπ-π < 2x-π/3 < 2kπ (k = 0, plus or minus 1, plus or minus 2 ...) and x is in the range of [0, π] (assign 0, 1 to k).

(2) From (1), we know that f(x)=2cos(2x-π/3)+m+ 1, and the maximum value on [0,6/π] is f(π/6)=m+3.

The minimum value is f(0)=m+2, and when x ∈ [0 0,6/π], -4.

So m+3 < 4, m+2 >-4, that is, the range of m is (-6, 1).

2.|a-b|=|(cosa-cosb,sina-sinb)|=√(cosa-cosb)^2+(sina-sinb)^2=√2-2(cosacosb+sinasinb)=√2-2cos(a-b)=2/5√5

cos(a-b)=？