(1)f'(x)=a+x-x²由题有f'(x)≥0对x∈(0,¾)恒成立,∴a≥(x²-x)max,在所给开区间上,x²-x≤¾²-¾=-3/16∴a≥-3/16
(2)由题欲证g(x)＞f'(x),令h(x)=-g(x)+f'(x)=x²-x+sinx+cosx-3/16,h'(x)=2x-1+cosx-sinx,h"(x)=2-sinx-cosx≥2-√2＞0,x∈(0,∞)∴h'(x)单增,而h'(0)=0,∴x>0时,h'(x)＞0,∴h(x)在定义域上单增,又∵h(0)>0,结合h单增可得结论