Đáp án:

$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1\\
a)Q = \dfrac{{3x - 3\sqrt x  - 3}}{{x + \sqrt x  - 2}} - \dfrac{{\sqrt x  + 1}}{{\sqrt x  + 2}} + \dfrac{{\sqrt x  - 2}}{{1 - \sqrt x }}\\
 = \dfrac{{3x - 3\sqrt x  - 3 - \left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right) - \left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{3x - 3\sqrt x  - 3 - \left( {x - 1} \right) - \left( {x - 4} \right)}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{3x - 3\sqrt x  - 3 - x + 1 - x + 4}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{x - 3\sqrt x  + 2}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 1} \right)}}\\
 = \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 2}}\\
c)Q = \dfrac{1}{3}\\
 \Leftrightarrow \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 2}} = \dfrac{1}{3}\\
 \Leftrightarrow 3\left( {\sqrt x  - 2} \right) = \sqrt x  + 2\\
 \Leftrightarrow 3\sqrt x  - 6 = \sqrt x  + 2\\
 \Leftrightarrow 2\sqrt x  = 8\\
 \Leftrightarrow \sqrt x  = 4\\
 \Leftrightarrow x = 16\left( {tmdk} \right)\\
Vậy\,x = 16\\
h)\left| Q \right| =  - Q\\
 \Leftrightarrow Q \le 0\\
 \Leftrightarrow \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 2}} \le 0\\
 \Leftrightarrow \sqrt x  - 2 \le 0\\
 \Leftrightarrow \sqrt x  \le 2\\
 \Leftrightarrow x \le 4\\
Vậy\,0 \le x \le 4;x \ne 1\\
k)Q \in Z\\
 \Leftrightarrow \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 2}} = \dfrac{{\sqrt x  + 2 - 4}}{{\sqrt x  + 2}}\\
 = 1 - \dfrac{4}{{\sqrt x  + 2}} \in Z\\
 \Leftrightarrow \left( {\sqrt x  + 2} \right) \in \left\{ {2;4} \right\}\\
 \Leftrightarrow \sqrt x  \in \left\{ {0;2} \right\}\\
 \Leftrightarrow x \in \left\{ {0;4} \right\}\left( {tmdk} \right)\\
m)Q = m\\
 \Leftrightarrow \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 2}} = m\\
 \Leftrightarrow \sqrt x  - 2 = m\sqrt x  + 2m\\
 \Leftrightarrow \left( {m - 1} \right).\sqrt x  =  - 2m - 2\\
 \Leftrightarrow \sqrt x  = \dfrac{{ - 2m - 2}}{{m - 1}}\left( {m \ne 1} \right)\\
\left\{ \begin{array}{l}
\sqrt x  \ge 0\\
\sqrt x  \ne 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{ - 2m - 2}}{{m - 1}} \ge 0\\
\dfrac{{ - 2m - 2}}{{m - 1}} \ne 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{2m + 2}}{{m - 1}} \le 0\\
 - 2m - 2 \ne m - 1
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
 - 1 \le m < 1\\
m \ne  - \dfrac{1}{3}
\end{array} \right.\\
Vậy\, - 1 \le m < 1;m \ne  - \dfrac{1}{3}
\end{array}$