1、f(1)=1+m/1=2,
m=1,
f(x)=x+1/x,
f(-x)=-x-1/x=-(x+1/x)=-f(x),
g(x)是奇函數(shù),
則g(-x)=-g(x),
F(-x)=f(-x)*g(-x)=[-f(x)]*[-g(x)]=f(x)*g(x)=F(x),
故F(x)是偶函數(shù).
2、f'(x)=1-1/x^2,
x>1==>x^2>1==>1/x^20,
故f(x)在(1,+∞）單調遞增.
若不用導數(shù),則用定義來判斷,
設在（1,+∞）區(qū)間內舍近求取二數(shù)x2>x1>1,
則f(x2)=x2+1/x2,
f(x1)=x1+1/x1,
f(x2)-f(x1)=(x2+1/x2)-(x1+1/x1)
=(x2-x1)+(1/x2-1/x1)
=(x2-x1)+(x1-x2)/(x1*x2)
=(x2-x1)[1-1/(x1x2)],
因x2>x1,則x2-x1>0,
x2>x1>1,則x1*x2>1,
1/(x1*x2)0,
則（x2-x1)[1-1/(x1x2)]>0,
f(x2)-f(x1)>0,
f(x2)>f(x1),
所以f(x)在（1,+∞）是單調增函數(shù).
3.f(1)=2