答：f(x)=sin²x+2√3sinxcosx+3cos²x
=1+√3sin2x+cos2x+1
=2*[(√3/2)sin2x+(1/2)cos2x]+2
=2sin(2x+π/6)+2
（1）f(a)=2sin(2a+π/6)+2=3
sin(2a+π/6)=1/2
因为：0
所以：a=π/3
（2）f(x)=2sin(2x+π/6)+2
0<=x<=π，π/6<=2x+π/6<=13π/6
所以：单调递增区间满足π/6<=2x+π/6<=π/2或者3π/2<=2x+π/6<=13π/6
所以：单调增区间为[0,π/3] ∪ [2π/3,π]
（3）f(x)=2sin(2x+π/6)+2>m-3在 [π/4,π/2]上恒成立
sin(2x+π/6)>(m-5)/2
π/4<=x<=π/2，2π/3<=2x+π/6<=7π/6
所以：sin(2x+π/6)>=sin(7π/6)=-1/2>(m-5)/3
解得：m<7/2
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024