Đáp án:

$2)-\dfrac{7}{12} \le m \le \dfrac{5}{12}\\ 3) 1 \le m \le \dfrac{4}{3}.$

Giải thích các bước giải:

$2)\\ y=x^3-3(2m+1)x^2+(12m+5)x+2\\ y'=3x^2-6(2m+1)x+12m+5$

Hàm số đồng biến trên $(-\infty;-1);[2;+\infty)$

$\Rightarrow y' \ge 0 \ \forall \ x \in(-\infty;-1) ; [2;+\infty)\\ \Leftrightarrow 3x^2-6(2m+1)x+12m+5 \ge 0 \ \forall \ x \in(-\infty;-1) ; [2;+\infty)\\ \Leftrightarrow 3x^2-12mx-6x+12m+5 \ge 0 \ \forall \ x \in(-\infty;-1) ; [2;+\infty)\\ \Leftrightarrow 12mx-12m \le 3x^2-6x+5 \ \forall \ x \in(-\infty;-1) ; [2;+\infty)\\ \Leftrightarrow 12m(x-1) \le 3x^2-6x+5 \ \forall \ x \in(-\infty;-1) ; [2;+\infty)(1)\\ \circledast x \in (-\infty;-1)\Rightarrow x-1<0\\ (1) \Leftrightarrow m \ge \dfrac{3x^2-6x+5}{12(x-1)} \ \forall \ x \in(-\infty;-1) \\ \Leftrightarrow m \ge g(x) \ \forall \ x \in(-\infty;-1) \\ \Leftrightarrow m \ge max_{g(x)} ; x \in(-\infty;-1) (*)\\ g(x)=\dfrac{3x^2-6x+5}{12(x-1)}\\ g'(x)=\dfrac{(3x^2-6x+5)'12(x-1)-(3x^2-6x+5)12(x-1)'}{(12(x-1))^2}\\ =\dfrac{(6x-6)(x-1)-(3x^2-6x+5)}{12(x-1)^2}\\ =\dfrac{3 x^2 - 6 x + 1}{12(x-1)^2}\\ g'(x)=0 \Leftrightarrow x=\dfrac{3\pm \sqrt{6}}{3}\\ BBT:$

\begin{array}{|c|ccccccccc|} \hline x&-\infty&&-1\\\hline y'&&+&\\\hline &&&-1\\y&&\nearrow&\\&-\infty\\\hline\end{array}

$\Rightarrow (*)\Leftrightarrow m \ge -\dfrac{7}{12}\\ \circledast x \in [2;+\infty) \Rightarrow x-1>0\\ (1) \Leftrightarrow m \le \dfrac{3x^2-6x+5}{12(x-1)} \ \forall \ x \in [2;+\infty)\\ \Leftrightarrow m \le g(x) \ \forall \ x \in [2;+\infty)\\ \Leftrightarrow m \le min_{g(x)} ; x \in [2;+\infty) (**)\\ BBT:$

\begin{array}{|c|ccccccccc|} \hline x&2&&+\infty\\\hline y'&&+&\\\hline &&&+\infty\\y&&\nearrow&\\&\dfrac{5}{12}\\\hline\end{array}

$\Rightarrow (**)\Leftrightarrow m \le \dfrac{5}{12}$

Kết hợp 2 trường hợp 

$\Rightarrow -\dfrac{7}{12} \le m \le \dfrac{5}{12}\\ 3)\\ y=\dfrac{1}{3}mx^3+2(m-1)x^2+(m-1)x+m\\ y'=mx^2+4(m-1)x+m-1$

Hàm số đồng biến trên $(-\infty;0);[2;+\infty)$

$\Rightarrow y' \ge 0 \ \forall \ x \in(-\infty;0); [2;+\infty)\\ \Leftrightarrow mx^2+4(m-1)x+m-1 \ge 0 \ \forall \ x \in(-\infty;0); [2;+\infty)\\ \Leftrightarrow mx^2+4mx-4x+m-1 \ge 0 \ \forall \ x \in(-\infty;0); [2;+\infty)\\ \Leftrightarrow mx^2+4mx+m\ge 4x+1 \ \forall \ x \in(-\infty;0); [2;+\infty)\\ \Leftrightarrow m(x^2+4x+1)\ge 4x+1 \ \forall \ x \in(-\infty;0); [2;+\infty) (1)$

$\circledast x^2+4x+1=0 \Leftrightarrow x=-2 \pm \sqrt{3}$

$(1) \Leftrightarrow 0 \ge -7 \pm 4\sqrt{3}$ luôn đúng

$\circledast x^2+4x+1>0 \Leftrightarrow x \in (-\infty; -2 -\sqrt{3}) \cup (-2 + \sqrt{3}; +\infty)$

Kết hợp với yêu cầu đề

$\Rightarrow x \in (-\infty; -2 -\sqrt{3}) \cup (-2 + \sqrt{3}; 0) \cup [2;+\infty)\\ (1) \Leftrightarrow  m \ge \dfrac{4x+1}{x^2+4x+1} \ \forall \ x \in (-\infty; -2 -\sqrt{3}) \cup (-2 + \sqrt{3}; 0) \cup [2;+\infty)\\ \Leftrightarrow  m \ge g(x) \ \forall \ x \in (-\infty; -2 -\sqrt{3}) \cup (-2 + \sqrt{3}; 0) \cup [2;+\infty)\\ \Leftrightarrow  m \ge max_{g(x)}; x \in (-\infty; -2 -\sqrt{3}) \cup (-2 + \sqrt{3}; 0) \cup [2;+\infty)(*)\\ g(x)= \dfrac{4x+1}{x^2+4x+1}\\ g'(x)= \dfrac{(4x+1)'(x^2+4x+1)-(4x+1)(x^2+4x+1)'}{(x^2+4x+1)^2}\\ =\dfrac{4(x^2+4x+1)-(4x+1)(2x+4)}{(x^2+4x+1)^2}\\ =\dfrac{-4 x^2 - 2 x}{(x^2+4x+1)^2}\\ =\dfrac{-2x(2x+1)}{(x^2+4x+1)^2}\\ g'(x) =0 \Leftrightarrow x=0; x=-\dfrac{1}{2}\\ BBT:$

\begin{array}{|c|ccccccccc|} \hline x&-\infty&&&-2 -\sqrt{3}&&-\dfrac{1}{2}&&-2 + \sqrt{3}&&&0&&&&2&&&\infty\\\hline y'&&-&&||&/&/&/&||&&+&0|&/&/&/&|&-&&\\\hline &0&&&||&/&/&/&||&&&1|&/&/&/&|\dfrac{9}{13}\\y&&\searrow&&||&/&/&/&||&&\nearrow&|&/&/&/&|&\searrow&\\&&&-\infty&||&/&/&/&||&-\infty&&|&/&/&/&|&&0\\\hline\end{array}

$\Rightarrow (*) \Leftrightarrow m \ge 1\\ \circledast x^2+4x+1<0 \Leftrightarrow  x \in (-2 -\sqrt{3};-2 + \sqrt{3})\\ (1) \Leftrightarrow  m \le \dfrac{4x+1}{x^2+4x+1} \ \forall \ x \in (-2 -\sqrt{3};-2 + \sqrt{3})\\ \Leftrightarrow m \le min_{g(x)};x \in (-2 -\sqrt{3};-2 + \sqrt{3})(**)\\ BBT$

\begin{array}{|c|ccccccccc|} \hline x&-2 -\sqrt{3}&&-\dfrac{1}{2}&&-2 +\sqrt{3}\\\hline y'&&-&0&+&\\\hline &+\infty&&&&+\infty\\y&&\searrow&&\nearrow&\\&&&\dfrac{4}{3}\\\hline\end{array}

$\Rightarrow (**) \Leftrightarrow m \le \dfrac{4}{3}$

Kết hợp 2 trường hợp 

$\Rightarrow 1 \le m \le \dfrac{4}{3}.$