Giải thích các bước giải:

a.Ta có: $y'=3x^2-3$
Xét $A\left(a, a^3-3a+2\right)$
      $A'\left(b, b^3-3b+3\right), \left(a\ne b\right)$
$\to AA'$ là tiếp tuyến của $\left(C\right)$ tại $A'$
$\to AA': y=\left(3b^2-3\right)\left(x-b\right)+b^3-3b+3$
Lại có: $A\in AA'$
$\to a^3-3a+2=\left(3b^2-3\right)\left(a-b\right)+b^3-3b+3$
$\to a^3-3a=\left(3b^2-3\right)\left(a-b\right)+b^3-3b$
$\to \left(a^3-b^3\right)-3\left(a-b\right)=\left(3b^2-3\right)\left(a-b\right)$
$\to \left(a-b\right)\left(a^2+ab+b^2\right)-3\left(a-b\right)=\left(3b^2-3\right)\left(a-b\right)$
$\to a^2+ab+b^2-3=3b^2-3$ vì $a\ne b$
$\to a^2+ab-2b^2=0$
$\to \left(a-b\right)\left(a+2b\right)=0$
$\to a+2b=0$
$\to a=-2b$
$\to A\left(-2b, -8b^3+6b+2\right), A'\left(b,b^3-3b+3\right)$
Để ko bị nhầm lẫn đặt
$A\left(-2a, -8a^3+6a+2\right), A'\left(a,a^3-3a+3\right)$
$B\left(-2b, -8b^3+6b+2\right), B'\left(b,b^3-3b+3\right)$
$C\left(-2c, -8c^3+6c+2\right), C'\left(c,c^3-3c+3\right)$
$\to \begin{cases}\vec{AB}=\left(2\left(a-b\right),8\left(a-b\right)\left(a^2+ab+b^2\right)-6\left(a-b\right)\right),\vec{A'B'}=\left(-\left(a-b\right), -\left(a-b\right)\left(b^2+ab+a^2-3\right)\right)\\\vec{AC}=\left(2\left(a-c\right),8\left(a-c\right)\left(a^2+ac+c^2\right)-6\left(a-c\right)\right),\vec{A'B'}=\left(-\left(a-c\right), -\left(a-c\right)\left(c^2+ac+a^2-3\right)\right)\end{cases}$
$\to \begin{cases}\vec{m_{AB}}=\left(1,4\left(a^2+ab+b^2\right)-3\right),\vec{m_{A'B'}}=\left(-1, -\left(b^2+ab+a^2-3\right)\right)\\\vec{m_{AC}}=\left(1,4\left(a^2+ac+c^2\right)-3\right),\vec{m_{A'B'}}=\left(-1, -\left(c^2+ac+a^2-3\right)\right)\end{cases}$
Với $\vec{m}$ là vector chỉ phương
Vì $A,B,C$ thẳng hàng
$\to \vec{m_{AB}}//\vec{m_{AC}}$
$\to \dfrac11=\dfrac{4\left(a^2+ab+b^2\right)-3}{4\left(a^2+ac+c^2\right)-3}$
$\to a^2+ab+b^2=a^2+ac+c^2$
$\to \dfrac{-1}{-1}=\dfrac{-\left(b^2+ab+a^2-3\right)}{-\left(c^2+ac+a^2-3\right)}$
$\to \vec{m_{A'B'}}//\vec{m_{A'C'}}$
b.Ta có: $\left(C\right): y=\dfrac{x+m}{x-2}$ có tiếp tuyến $\to m\ne -2$
Gọi $B\left(b,\dfrac{b+m}{b-2}\right), \left(c,\dfrac{c+m}{c-2}\right), \left(b\ne c\ne 2\right)$
Ta có: $y'=-\dfrac{m+2}{\left(x-2\right)^2}$
Vì $AB,AC$ là tiếp tuyến của $\left(C\right)$ tại $B,C$
$\to$Phương trình $AB,AC$ là:
$\begin{cases}y=-\dfrac{m+2}{\left(b-2\right)^2}\left(x-b\right)+\dfrac{b+m}{b-2}\\ y=-\dfrac{m+2}{\left(c-2\right)^2}\left(x-c\right)+\dfrac{c+m}{c-2}\end{cases}$
Mà $AB,AC$ đi qua $A$
$\to -\dfrac{m+2}{\left(b-2\right)^2}\left(1-b\right)+\dfrac{b+m}{b-2}=-\dfrac{m+2}{\left(c-2\right)^2}\left(1-c\right)+\dfrac{c+m}{c-2}$
$\to \dfrac{\left(m+2\right)\left(b-1\right)}{\left(b-2\right)^2}+\dfrac{b-2+m+2}{b-2}=\dfrac{\left(m+2\right)\left(c-1\right)}{\left(c-2\right)^2}+\dfrac{c-2+m+2}{c-2}$

$\to \dfrac{\left(m+2\right)\left(b-2+1\right)}{\left(b-2\right)^2}+\dfrac{m+2}{b-2}+1=\dfrac{\left(m+2\right)\left(c-2+1\right)}{\left(c-2\right)^2}+\dfrac{m+2}{c-2}+1$
$\to \dfrac{m+2}{b-2}+\dfrac{m+2}{\left(b-2\right)^2}+\dfrac{m+2}{b-2}=\dfrac{m+2}{c-2}+\dfrac{m+2}{\left(c-2\right)^2}+\dfrac{m+2}{c-2}$
$\to \dfrac{1}{b-2}+\dfrac{1}{\left(b-2\right)^2}+\dfrac{1}{b-2}=\dfrac{1}{c-2}+\dfrac{1}{\left(c-2\right)^2}+\dfrac{1}{c-2}$
$\to \dfrac{1}{\left(b-2\right)^2}+\dfrac{2}{b-2}=\dfrac{1}{\left(c-2\right)^2}+\dfrac{2}{c-2}$
$\to \left(\dfrac{1}{\left(b-2\right)^2}-\dfrac{1}{\left(c-2\right)^2}\right)+\left(\dfrac{2}{b-2}-\dfrac{2}{c-2}\right)=0$
$\to \left(\dfrac{1}{b-2}-\dfrac{1}{c-2}\right)\left(\dfrac{1}{b-2}+\dfrac{1}{c-2}\right)+2\left(\dfrac{1}{b-2}-\dfrac{1}{c-2}\right)=0$
$\to \dfrac{1}{b-2}+\dfrac{1}{c-2}+2=0$ vì $b\ne c$
$\to\dfrac{1}{b-2}+\dfrac{1}{c-2}=-2\left(*\right)$

Lại có: $\Delta ABC$ đều 
$\to AB^2=AC^2$
$\to \left(b-1\right)^2+\left(\dfrac{b+m}{b-2}-2\right)^2=\left(c-1\right)^2+\left(\dfrac{c+m}{c-2}-2\right)^2$
$\to \left(\left(b-1\right)^2-\left(c-1\right)^2\right)+\left(\left(\dfrac{b+m}{b-2}-2\right)^2-\left(\dfrac{c+m}{c-2}-2\right)^2\right)=0$
$\to \left(b-c\right)\left(b+c-2\right)+\left(\dfrac{b+m}{b-2}-\dfrac{c+m}{c-2}\right)\left(\dfrac{b+m}{b-2}+\dfrac{c+m}{c-2}-4\right)=0$
$\to \left(b-c\right)\left(b+c-2\right)+\left(\left(\dfrac{m+2}{b-2}+1\right)-\left(\dfrac{m+2}{c-2}+1\right)\right)\left(\left(\dfrac{m+2}{b-2}+1\right)+\left(\dfrac{m+2}{c-2}+1\right)-4\right)=0$
$\to \left(b-c\right)\left(b+c-2\right)+\left(\dfrac{m+2}{b-2}-\dfrac{m+2}{c-2}\right)\left(\dfrac{m+2}{b-2}+\dfrac{m+2}{c-2}-2\right)=0$
$\to \left(b-c\right)\left(b+c-2\right)+\left(m+2\right)\left(\dfrac{1}{b-2}-\dfrac{1}{c-2}\right)\left(\left(m+2\right)\left(\dfrac{1}{b-2}+\dfrac{1}{c-2}\right)-2\right)=0$
$\to \left(b-c\right)\left(b+c-2\right)+\left(m+2\right)\cdot\dfrac{-\left(b-c\right)}{\left(b-2\right)\left(c-2\right)}\cdot\left(\left(m+2\right)\cdot\left(-2\right)-2\right)=0$
$\to \left(b+c-2\right)-\dfrac{\left(m+2\right)}{\left(b-2\right)\left(c-2\right)}\cdot\left(-2m-6\right)=0$
$\to \left(b+c-2\right)+\dfrac{2\left(m+3\right)\left(m+2\right)}{\left(b-2\right)\left(c-2\right)}=0$
$\to \left(b+c-2\right)\left(b-2\right)\left(c-2\right)+2\left(m+3\right)\left(m+2\right)=0$
$\to \left(b+c-2\right)\left(b-2\right)\left(c-2\right)=-2\left(m+3\right)\left(m+2\right)$
$\to \left(b+c-2\right)\cdot -2\left(b-2\right)\left(c-2\right)=4\left(m+3\right)\left(m+2\right)$
$\to \left(b+c-2\right)\cdot \left(b+c-4\right)=4\left(m+3\right)\left(m+2\right)$ vì $\left(*\right)$
$\to \left(b+c\right)^2-6\left(b+c\right)+8=4\left(m+2\right)^2+4\left(m+2\right)$
$\to \left(b+c\right)^2-6\left(b+c\right)+9=4\left(m+2\right)^2+4\left(m+2\right)+1$
$\to \left(b+c-3\right)^2=\left(2\left(m+2\right)+1\right)^2$
$\to \left(b+c-3\right)^2=\left(2m+5\right)^2$
$\to b+c-3=2m+5\to b+c-4=2m+4\to \dfrac{b+c-4}{-2}=-(m+2)\to \left(b-2\right)\left(c-2\right)=-(m+2)$
Hoặc $b+c-3=-2m-5\to b+c-4=-2m-6\to \dfrac{b+c-4}{-2}=m+3\to \left(b-2\right)\left(c-2\right)=m+3$


Ta có: $\Delta ABC$ đều $\to \widehat{BAC}=60^o$
$\to \tan\widehat{BAC}=\sqrt{3}$
Ta có: $k_1=-\dfrac{m+2}{\left(b-2\right)^2}, k_2=-\dfrac{m+2}{\left(c-2\right)^2}$ lần lượt là hệ số góc của $AB,AC$ với $Ox$
Chia góc $\widehat{BAC}$ bằng đường thẳng đi qua $A$ và song song với $Ox$ thành hai góc $\alpha,\beta$
$\to \tan\alpha= -\dfrac{m+2}{\left(b-2\right)^2}, \tan\beta=-\dfrac{m+2}{\left(c-2\right)^2}$
$\to 60^o=\alpha+\beta$
$\to\tan60^o=\tan\left(\alpha+\beta\right)$
$\to\sqrt{3}=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
$\to\sqrt{3}=\dfrac{-\dfrac{m+2}{\left(b-2\right)^2}-\dfrac{m+2}{\left(c-2\right)^2}}{1-\dfrac{m+2}{\left(b-2\right)^2}\cdot\dfrac{m+2}{\left(c-2\right)^2}}$
$\to\sqrt{3}=\dfrac{-\left(m+2\right)\left(\dfrac{1}{\left(b-2\right)^2}+\dfrac{1}{\left(c-2\right)^2}\right)}{1-\left(m+2\right)^2\cdot\dfrac{1}{\left(b-2\right)^2}\cdot\dfrac{1}{\left(c-2\right)^2}}$
$\to\sqrt{3}=\dfrac{-\left(m+2\right)\left(\left(\dfrac{1}{b-2}+\dfrac{1}{c-2}\right)^2-2\cdot\dfrac{1}{b-2}\cdot\dfrac{1}{c-2}\right)}{1-\left(m+2\right)^2\cdot\dfrac{1}{\left(b-2\right)^2}\cdot\dfrac{1}{\left(c-2\right)^2}}$ $\to\sqrt{3}=\dfrac{-\left(m+2\right)\left(\left(-2\right)^2-\dfrac{2}{\left(b-2\right)\left(c-2\right)}\right)}{1-\left(m+2\right)^2\cdot\dfrac{1}{\left(b-2\right)^2}\cdot\dfrac{1}{\left(c-2\right)^2}}$
$\to\sqrt{3}=\dfrac{\left(m+2\right)\left(4-\dfrac{2}{\left(b-2\right)\left(c-2\right)}\right)}{1-\left(m+2\right)^2\cdot\dfrac{1}{\left(b-2\right)^2}\cdot\dfrac{1}{\left(c-2\right)^2}}$

$\to 1-\left(m+2\right)^2\cdot \dfrac{1}{\left(b-2\right)^2}\cdot\dfrac{1}{\left(c-2\right)^2}\ne0$

$\to \left(b-2\right)^2\left(c-2\right)^2\ne \left(m+2\right)^2$

$\to \left(b-2\right)\left(c-2\right)\ne \pm\left(m+2\right)$

$\to \left(b-2\right)\left(c-2\right)=m+3$

$\to\sqrt{3}=\dfrac{\left(m+2\right)\left(4-\dfrac{2}{m+3}\right)}{1-\left(m+2\right)^2\cdot\dfrac{1}{(m+3)^2}}$

$\to m=\dfrac{-5\pm\sqrt{1+2\sqrt{3}}}{2}$