f[x]=(1+cos2x)/2+1/2·sin2x
=1/2+1/2(cos2x+sin2x)
=1/2+√2/2·sin(2x+π/4)
(1)f(3π/8)=1/2+√2/2·sin(3π/4+π/4)
=1/2.
（2）又2kπ-π/2≤2x+π/4≤2kπ+π/2
解得
kπ-3π/8≤x≤kπ+π/8,
∴f(x)的单调递增区间是[kπ-3π/8,kπ+π/8].