Đáp án:

$\begin{array}{l}
a)2 \le \left| {x + 1} \right|\\
 \Rightarrow \left[ \begin{array}{l}
x + 1 \ge 2\\
x + 1 \le  - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x \ge 1\\
x \le  - 3
\end{array} \right.\\
b)\left| {x - 2} \right| < 3x - 2\\
 \Rightarrow \left\{ \begin{array}{l}
3x - 2 > 0\\
 - 3x + 2 < x - 2 < 3x - 2
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x > \dfrac{2}{3}\\
1 < x
\end{array} \right.\\
 \Rightarrow x > 1\\
c)\left( {2x + 1} \right)\left( {{x^2} + x - 30} \right) > 0\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x + 1 > 0\\
{x^2} + x - 30 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x + 1 < 0\\
{x^2} + x - 30 < 0
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x >  - \dfrac{1}{2}\\
\left( {x - 5} \right)\left( {x + 6} \right) > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x <  - \dfrac{1}{2}\\
\left( {x - 5} \right)\left( {x + 6} \right) < 0
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x >  - \dfrac{1}{2}\\
x > 5/x <  - 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x <  - \dfrac{1}{2}\\
 - 6 < x < 5
\end{array} \right.
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x > 5\\
 - 6 < x <  - \dfrac{1}{2}
\end{array} \right.\\
2)\dfrac{{x - 2}}{{x + 1}} > \dfrac{x}{{x + 3}}\\
 \Rightarrow \dfrac{{x + 1 - 3}}{{x + 1}} > \dfrac{{x + 3 - 3}}{{x + 3}}\\
 \Rightarrow 1 - \dfrac{3}{{x + 1}} > 1 - \dfrac{3}{{x + 3}}\\
 \Rightarrow \dfrac{3}{{x + 1}} < \dfrac{3}{{x + 3}}\\
 \Rightarrow \dfrac{1}{{x + 1}} < \dfrac{1}{{x + 3}}\\
 \Rightarrow \dfrac{{x + 3 - x - 1}}{{\left( {x + 1} \right)\left( {x + 3} \right)}} < 0\\
 \Rightarrow \dfrac{2}{{\left( {x + 1} \right)\left( {x + 3} \right)}} < 0\\
 \Rightarrow \left( {x + 1} \right)\left( {x + 3} \right) < 0\\
 \Rightarrow  - 3 < x <  - 1
\end{array}$