这道数学题怎么做？求过程。谢谢了。

f(x)=(1+cosx)(sinx)^2-2sin(x+π/4)sin(x-π/4)
=(1/2)(1+cosx)(1-cos2x)+cos2x-cos(π/2）
=（1/2）(1+cosx+cos2x-cosxcos2x),
(1)tana=2,
∴cosa=土1/√5，cos2a=-3/5,
∴f(a)=(1/2)[1-3/5土(1+3/5)/√5]
=(5土4√5)/25.
(2)f(x)=(cosx)^2+cosx[1-(cosx)^2],
x∈[π/12，π/2],
u=cosx的值域是[0,(√6+√2)/4],
f(x)=u+u^2-u^3,记为g(u),
g'(u)=1+2u-3u^2=(1-u)(1+3u)>0,
∴g(u)是增函数，g(0)=0,
g[(√6+√2)/4]=(√6+√2)/4*[1+(√6+√2)/4-(2+√3)/4]
=(√6+√2)(2+√6+√2-√3)/16
=(2√6+6+2√3-3√2+2√2+2√3+2-√6)/16
=(8+√6+4√3-√2)/16.
∴f(x)的取值范围是[0,(8+√6+4√3-√2)/16].