(1) 若函数f(x)=sinx/2cosx/2+cos2x/2-1/2 => 化简得f(x)=1/2*sinx+1/2*cos2x-1/2
=1/2sinx+1/2*[1-2sin^2(x)]-1/2
=-sin^2(x)+1/2sinx
当f(α)=√2/4时,有 -sin^2(x)+1/2sinx=√2/4 => -(sinx-1/4)^2=√2/4-1/16>0
但是左边 √2sin(x+π/4)-2=√2 => √2 * sin(x+π/4)=√2+2
sin(x+π/4)=1+√2>1 因为|sin(x+π/4)|《1,所以在（0,2π）上,无α满足上面条件.
（2）当x 在[-π/4,π]上时 由 -π/4《x《π => 0《x《5/4π
所以 -1《√2 * sin(x+π/4)《√2 => -1/4《√2 /4 sin(x+π/4)《√2 /4
=> -1/4-1/2《√2/4 sin(x+π/4)-1/2《√2/4-1/2
=> f(x)最大值√2/4-1/2 ;最小值-3/4