(1)当m=0时,函数f(x)=(1+
1
tanx )sin 2x=
sinx+cosx
sinx •sin 2x=sin 2x+sinxcosx=
1-cos2x
2 +
1
2 sin2x=
1
2 +
2
2 sin(2x-
π
4 ).
∵
π
8 ≤x≤
3π
4 ,∴0≤2x-
π
4 ≤
5π
4 ,∴-
2
2 ≤sin(2x-
π
4 )≤1,0≤f(x)≤
1+
2
2 ,
故f(x)在区间[
π
8 ,
3π
4 ]上的取值范围为[0
1+
2
2 ,].
(2)∵当tana=2时,f(a)=
3
5 ,∴sin 2a=
4
5 ,cos 2a=
1
5 .
再由f(a)=(1+
1
tana )sin 2a+msin(a+
π
4 )sin(a-
π
4 )=
3
2 sin 2a+m(
1
2 sin 2a-
1
2 cos 2a )=
12+3m
10 ,
可得
12+3m
10 =
3
5 ,解得m=-2.