设x∈R,函数f(x)=cos(wx+f)(w>0,-π/2

(1)解析：∵函数f(x)=cos(wx+f)(w>0,-π/2<f<0)的最小正周期为π∴w=2π/π=2,f(x)=cos(2x+f)∵f(π/4)=√3/2f(π/4)=cos(π/2+f)=-sin(f)=√3/2==>f=-π/3∴w=2,f=-π/3(2)解析：(3)解析：∵f(x)> √2/2f(x)=cos(2x-π/3)>√2/2√2/2<cos(2x-π/3)<=12kπ-π/4<2x-π/3<2kπ+π/4==>kπ+π/24<x<kπ+7π/24 （k为整数）∴当kπ+π/24<x<kπ+7π/24时,f(x)> √2/2