(1)f(x)=sin2x-2cos²x+m
=sin2x-(1+cos2x)+m
=√2sin(2x-π/4)+m-1
f(x)=√2sin(2x-π/4)+m-1,经过点(π/8,0)
√2sin(2*π/8-π/2)+m-1=0
m=1
f(x)=√2sin(2x-π/4),最大值=√2
(2)f(α/2)=)=√2sin(α-π/4)=3√2/5
sin(α-π/4)=3/5
α∈(0,π/2),α-π/4∈(-π/4,π/4),
cos(α-π/4)=4/5
sinα=sin[(α-π/4)+π/4]=√2/2[sin(α-π/4)+cos(α-π/4)]
=√2/2(3/5+4/5)
=7√2/10