答：
f(x)=2(sinx)^4+2(cosx)^4+(cos2x)^2-3
=2*[(sinx)^2+(cosx)^2]^2-4(sinxcosx)^2+(cos2x)^2-3
=2-(sin2x)^2+(cos2x)^2-3
=cos4x-1
所以：
f(x)的最小正周期T=2π/4=π/2
π/16<=x<=3π/16,π/4<=4x<=3π/4
所以：-√2/2<=cos4x<=√2/2
所以：f(x)的最小值为-1-√2/2