f(π/4)= m( 1+sinπ/2)+cosπ/2

=m( 1+ 1)+0

=2m=2

So m= 1.

(2)f(x)= m( 1+sin2x)+cos2x = f(x)= sin2x+cos2x+ 1

=√2sin(2x+π/4)+ 1

When 2x+π/4=2kπ+3/2π, that is, x=kπ+5/8π, the minimum value is 1-√2.

That is, the x set of the minimum value is x=kπ+5/8π (k belongs to r).