1: Given that the maximum value of F (x) = a-bcosx is 5 / 2 and the minimum value is negative 1 / 2, we can find the maximum value and the minimum positive period of G (x) = - 4A sinbx 2: Given that sin (a + π) = 3 / 5 and Sina × cosa < 0, the value of {2Sin (a - π) + 3tan (3 π - a)} / {4cos (A-3 π)} can be obtained

1: Given that the maximum value of F (x) = a-bcosx is 5 / 2 and the minimum value is negative 1 / 2, we can find the maximum value and the minimum positive period of G (x) = - 4A sinbx 2: Given that sin (a + π) = 3 / 5 and Sina × cosa < 0, the value of {2Sin (a - π) + 3tan (3 π - a)} / {4cos (A-3 π)} can be obtained

1. Since f (x) = a-bcosx, Fmax = 5 / 2, Fmin = 1 / 2,
Then, a + B = 5 / 2, A-B = 1 / 2, so, a = 3 / 2, B = 1
So, G (x) = - 6sinx
The minimum positive period is 2 π
2. From sin (a + π) = 3 / 5, Sina = - 3 / 5, and Sina × cosa < 0,
Then, a is the angle of the fourth quadrant,
｛2sin（a+π）+3tan（3π-a）｝/｛4cos（a-3π）｝
=(-2sina-3tana)/-4cosa
=(3/5*2-3*3/4)/（-4*4/5）
=21/64