Đáp án:

$\begin{array}{l}
1)Dkxd:\left\{ \begin{array}{l}
a \ge 0\\
a \ne 0\\
\sqrt a  \ne 1\\
\sqrt a  \ne 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
a \ne 1\\
a \ne 4
\end{array} \right.\\
Vậy\,a > 0;a \ne 1;a \ne 4\\
2)\\
A = \left( {\dfrac{1}{{\sqrt a  - 1}} - \dfrac{1}{{\sqrt a }}} \right):\left( {\dfrac{{\sqrt a  + 1}}{{\sqrt a  - 2}} - \dfrac{{\sqrt a  + 2}}{{\sqrt a  - 1}}} \right)\\
 = \dfrac{{\sqrt a  - \left( {\sqrt a  - 1} \right)}}{{\sqrt a \left( {\sqrt a  - 1} \right)}}:\dfrac{{\left( {\sqrt a  + 1} \right)\left( {\sqrt a  - 1} \right) - \left( {\sqrt a  + 2} \right)\left( {\sqrt a  - 2} \right)}}{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  - 1} \right)}}\\
 = \dfrac{1}{{\sqrt a \left( {\sqrt a  - 1} \right)}}.\dfrac{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  - 1} \right)}}{{a - 1 - a + 4}}\\
 = \dfrac{1}{{\sqrt a }}.\dfrac{{\sqrt a  - 2}}{3}\\
 = \dfrac{{\sqrt a  - 2}}{{3\sqrt a }}\\
2)x = 7 - 4\sqrt 3 \left( {tmdk} \right)\\
 = 4 - 2.2\sqrt 3  + 3\\
 = {\left( {2 - \sqrt 3 } \right)^2}\\
 \Leftrightarrow \sqrt x  = 2 - \sqrt 3 \\
 \Leftrightarrow A = \dfrac{{\sqrt a  - 2}}{{3\sqrt a }} = \dfrac{{2 - \sqrt 3  - 2}}{{3\left( {2 - \sqrt 3 } \right)}}\\
 = \dfrac{{ - \sqrt 3 }}{{3\left( {2 - \sqrt 3 } \right)}}\\
 = \dfrac{1}{{\sqrt 3 \left( {\sqrt 3  - 2} \right)}}\\
 = \dfrac{1}{{3 - 2\sqrt 3 }}\\
 = \dfrac{{3 + 2\sqrt 3 }}{{{3^2} - {{\left( {2\sqrt 3 } \right)}^2}}}\\
 = \dfrac{{ - 3 - 2\sqrt 3 }}{3}\\
3)A = \dfrac{1}{4}\\
 \Leftrightarrow \dfrac{{\sqrt a  - 2}}{{3\sqrt a }} = \dfrac{1}{4}\\
 \Leftrightarrow 4\sqrt a  - 8 = 3\sqrt a \\
 \Leftrightarrow \sqrt a  = 8\\
 \Leftrightarrow a = 64\left( {tmdk} \right)\\
Vậy\,a = 64\\
4)A > 0\\
 \Leftrightarrow \dfrac{{\sqrt a  - 2}}{{3\sqrt a }} > 0\\
 \Leftrightarrow \sqrt a  - 2 > 0\\
 \Leftrightarrow \sqrt a  > 2\\
 \Leftrightarrow a > 4\\
Vậy\,a > 4
\end{array}$