(1）关于x的不等式f(x)=x^2-4mx+12m<=0在[-3,-1]上恒成立,
<==>f(-3)=9+24m<=0,且f(-1)=1+16m<=0,
解得m<=-3/8,且m<=-1/16,
∴m<=-3/8,为所求.
(2)关于x的不等式f(x)=x^2-4mx+12m>=0在[-3,-1]上恒成立,
分3种情况:
i)2m∈[-3,-1],f(2m)=-4m^2+12m>=0,
即m∈[-3/2,-1/2],m^2-3m<=0,0<=m<=3,不可能.
ii)m<-3/2,f(-3)=9+24m>=0,m>=-3/8,也不可能.
iii)m>-1/2,f(-1)=1+16m>=0,m>=-1/16,
∴m>=-1/16,为所求.