(1) m=0时
f(x)=sin^2x+sinxcosx
=(1/2)[1-cos2x+sin2x]
=(√2/2)sin(2x-π/4)+1/2
当2x-π/4=π/2 x=3π/8是f(x)max=(√2+1)/2
而f(π/8)=(√2/2)*0+1/2=1/2 f(3π/4)=(√2/2)sin(5π/4)+1/2=1/2-1/2=0
∴f（x）在区间[ π/8 ,3π/4 ]上的取值范围是[0,(√2+1)/2]
(2) 当tanα=2时,cotα=1/2 sin^2α=1/(1+cot^2α)=4/5 cos^2α=3/5
此时f(α)=3/5
∴3/5=(1+1/2)*(4/5)+m*(1/2)(sin^2α-cos^2α)=6/5+m*(1/2)(4/5-3/5)
∴m/10=3/5-6/5
∴m=-6(1/2)[1-cos2x+sin2x]=(√2/2)sin(2x-π/4)+1/2请赐教(*^__^*)(1/2)[1-cos2x+sin2x]=(1/2)(sin2x-cos2x)+1/2=(√2/2)[(√2/2)sin2x-(√2/2)cos2x]+1/2==(√2/2)sin(2x-π/4)+1/2