So f =1-sin2x-√ 3cos2x =1-2sin (2x+pi/3).

If the center of symmetry is 2x+PI/3=0, there are many points (-pi/6+k pie, 0).

Monotone interval subtraction makes 2 KPI-pi/2

Similar increasing interval

(2) equivalent to m-2

That is, find the value range of the function in the interval [Pi/4, Pi/2].

There is a single decrease in [pi/4, pi/3] and a monotonic increase in pi/3, pi/2.

The minimum value is f(pi/3)= 1, and the maximum value is f(pi/2)= 1+(√3)/2.

That is m-2 < 1 and m+2 >; 1+(√3)/2

Then (√ 3)/2-1< m <; three

Hope to adopt, thank you.