$
\left\{\begin{matrix}
6x^2 + x - 2 > 0 \ \ (1) \\
-x^2 + 6x + m^2 - 9 < 0 \ (2)
\end{matrix}\right.$
+ $(1) \Leftrightarrow 6 \left ( x - \dfrac{1}{2} \right) \left ( x + \dfrac{2}{3} \right) > 0
\Leftrightarrow \left[\begin{matrix}
x > \dfrac{1}{2} \\
\\
x < - \dfrac{2}{3}
\end{matrix}\right.
$
$\Leftrightarrow S_1 = \left ( - \infty; - \dfrac{2}{3} \right) \cup \left ( \dfrac{1}{2} ; + \infty \right)$
+ $(2) \ -x^2 + 6x + m^2 - 9 < 0 \\
\Leftrightarrow x^2 - 6x + 9 - m^2 > 0 \\
\Leftrightarrow (x-3)^2 -m^2 > 0 \\
\Leftrightarrow (x-3+m)(x-3-m) > 0 \\
\Leftrightarrow \left[\begin{matrix}
\left\{\begin{matrix}
x - 3 - m > 0 \\
x - 3 + m > 0
\end{matrix}\right. \\
\\
\left\{\begin{matrix}
x - 3 - m < 0 \\
x - 3 + m < 0
\end{matrix}\right.
\end{matrix}\right. \\
\Leftrightarrow \left[\begin{matrix}
\left\{\begin{matrix}
x > m+3 \\
x > -m+3
\end{matrix}\right. \\
\\
\left\{\begin{matrix}
x < m+3 \\
x < -m+3
\end{matrix}\right.
\end{matrix}\right. \\
\Leftrightarrow
\left[ \begin{matrix}
x > m+3 \\
x < -m+3
\end{matrix}\right.
$
$S_2 = (- \infty ; -m+3) \cup (m+3; + \infty)$
Hệ bpt có nghiệm $\Leftrightarrow$ xảy ra 3 trường hợp
+ TH1: $m+3 < - \dfrac{2}{3} \Leftrightarrow m < - \dfrac{11}{3}$
+ TH2: $
\left\{ \begin{matrix}
-m+3 < - \dfrac{2}{3} \\
m+3 > \dfrac{1}{2}
\end{matrix}\right. \\
\Leftrightarrow
\left\{ \begin{matrix}
-m < - \dfrac{11}{3} \\
m > - \dfrac{5}{2}
\end{matrix}\right. \\
\Leftrightarrow
\left\{ \begin{matrix}
m > \dfrac{11}{3} \\
m > - \dfrac{5}{2}
\end{matrix}\right. \\
\Leftrightarrow m > \dfrac{11}{3}
$
+TH3: $-m + 3 > \dfrac{1}{2} \Leftrightarrow -m > - \dfrac{5}{2} \Leftrightarrow m < \dfrac{5}{2}$