[1]∵f(x)=-4x+a为单调减函数∴f(x)在x∈(-1,2)的值为(a-4,a+8)∵|f(x)|＜6解集为(-1,2)∴a-4≥-6,或者a+8≤6,解得 a=-2[2]∵f(x)=-4x-2∴(4x+m)/f(x)＞0即(4x+m)/(4x+2)＜0当m=2时无解.当m＞2时,4x+m＞4x+2,即4x+2...