(1) 将（π/2,1)代入 原解析式,得 1=m·sin(π/2) +cos(π/2) ,m=1
则 f(x)=sinx+cosx
=√2 * sin(x +π/4)
w=1,最小正周期 T= 2π/w =2π
(2)因为 f(x) =√2 * sin(x +π/4)
f(a+π/4)=3√2/5
所以 √2 *sin(a+π/2)=3√2/5
sin(a+π/2)=3/5
cos(a)=3/5 则,sin(a)=4/5
则 f(2a-π/4) = √2 *sin(2a) = √2*2sin(a)cos(a)
= 24√2 / 25