f(x)=sin^2x+mcosx-1=1-cos^2x+mcosx-1=-cos^2x+mcosx=-(cosx-m/2)^2+m^2/4令t=cosx,则f(t)=-(t-m/2)^2+m^2/4=-t^2+mt∵t=cosx∈[-1,1]∴当t=m/2∈[-1,1],即m∈[-2,2]时,最大值g(m)=m^2/4；当t=m/2∈(-∞,-1],即m∈(...