解答：
∵ π/3<α<3π/4
∴ π/12<α-π/4<π/2
∵ sin(α-π/4)=3/5
∴ cos(α-π/4)=√[1-sin²(α-π/4)]=√(1-9/25)=4/5
∴ sinα=sin[(α-π/4)+π/4]
=sin(α-π/4)cos(π/4)+cos(α-π/4)sin(π/4)
=(3/5)*(√2/2)+(4/5)*(√2/2)
=7√2/10
cosα=cos[(α-π/4)+π/4]
=cos(α-π/4)cos(π/4)-sin(α-π/4)sin(π/4)
=(4/5)*(√2/2)-(3/5)*(√2/2)
=√2/10
∴(sinα-cos2α+1)/tanα
=(sinα+2sin²α)/(sinα/cosα)
=sinα(1+2sinα)/(sinα/cosα)
=cosα(1+2sinα)
=（√2/10）*（1+7√2/5）
=（5√2+14）/50
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