Đáp án:

$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
a > 0\\
a\# 1
\end{array} \right.\\
P = \left( {\dfrac{{a + 3\sqrt a  + 2}}{{\left( {\sqrt a  + 2} \right)\left( {\sqrt a  - 1} \right)}} + \dfrac{{a + \sqrt a }}{{a - 1}}} \right)\\
:\left( {\dfrac{1}{{\sqrt a  + 1}} + \dfrac{1}{{\sqrt a  - 1}}} \right)\\
 = \left( {\dfrac{{\left( {\sqrt a  + 1} \right)\left( {\sqrt a  + 2} \right)}}{{\left( {\sqrt a  + 2} \right)\left( {\sqrt a  - 1} \right)}} - \dfrac{{\sqrt a \left( {\sqrt a  + 1} \right)}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}} \right)\\
:\dfrac{{\sqrt a  - 1 + \sqrt a  + 1}}{{\left( {\sqrt a  + 1} \right)\left( {\sqrt a  - 1} \right)}}\\
 = \left( {\dfrac{{\sqrt a  + 1}}{{\sqrt a  - 1}} - \dfrac{{\sqrt a }}{{\sqrt a  - 1}}} \right).\dfrac{{\left( {\sqrt a  + 1} \right)\left( {\sqrt a  - 1} \right)}}{{2\sqrt a }}\\
 = \dfrac{1}{{\sqrt a  - 1}}.\dfrac{{\left( {\sqrt a  + 1} \right)\left( {\sqrt a  - 1} \right)}}{{2\sqrt a }}\\
 = \dfrac{{\sqrt a  + 1}}{{2\sqrt a }}\\
b)P =  - 2\\
 \Leftrightarrow \dfrac{{\sqrt a  + 1}}{{2\sqrt a }} =  - 2\\
 \Leftrightarrow \sqrt a  + 1 =  - 4\sqrt a \\
 \Leftrightarrow 5\sqrt a  + 1 = 0\left( {ktm} \right)\\
Vậy\,a \in \emptyset \\
c)P \le \left| { - 2} \right|\\
 \Leftrightarrow P \le 2\\
 \Leftrightarrow \dfrac{{\sqrt a  + 1}}{{2\sqrt a }} \le 2\\
 \Leftrightarrow \dfrac{{\sqrt a  + 1 - 4\sqrt a }}{{2\sqrt a }} \le 0\\
 \Leftrightarrow \dfrac{{1 - 3\sqrt a }}{{2\sqrt a }} \le 0\\
 \Leftrightarrow 1 - 3\sqrt a  \le 0\\
 \Leftrightarrow \sqrt a  \ge \dfrac{1}{3}\\
 \Leftrightarrow a \ge \dfrac{1}{9}\\
Vậy\,a \ge \dfrac{1}{9};a\# 1\\
d)a = \sqrt {13 + 2\sqrt {12} } \\
 = \sqrt {12 + 2\sqrt {12}  + 1} \\
 = \sqrt {{{\left( {\sqrt {12}  + 1} \right)}^2}} \\
 = \sqrt {12}  + 1 = 2\sqrt 3  + 1\\
 \Leftrightarrow P = \dfrac{{\sqrt a  + 1}}{{2\sqrt a }}\\
 = \dfrac{{\sqrt {2\sqrt 3  + 1}  + 1}}{{2\sqrt {2\sqrt 3  + 1} }} = 0,736
\end{array}$