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f(x)=2sinxcosx+cos2x
=sin2x+cos2x
=√2sin(2x+π/4)
(1)
f(π/4)=√2sin(π/2+π/4)=√2*sin(3π/4)=√2*√2/2=1
(2)
a∈(0,π)
a/2∈(0,π/2)
f(a/2)=√2sin(a+π/4)=√2/2
∴sin(a+π/4)=1/2
∵a∈(0,π)
∴a+π/4=5π/6
∴a=5π/6-π/4
sina
=sin(5π/6-π/4)
=sin5π/6cosπ/4-cos5π/6*sinπ/4
=1/2*√2/2+√3/2*√2/2
=(√2+√6)/2