Đáp án:

$\begin{array}{l}
1)a \ge 0;a \ne 9;a \ne 4\\
A = \dfrac{{2\sqrt a  - 9}}{{a - 5\sqrt a  + 6}} - \dfrac{{\sqrt a  + 3}}{{\sqrt a  - 2}} - \dfrac{{2\sqrt a  + 1}}{{3 - \sqrt a }}\\
 = \dfrac{{2\sqrt a  - 9}}{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  - 3} \right)}} - \dfrac{{\sqrt a  + 3}}{{\sqrt a  - 2}} + \dfrac{{2\sqrt a  + 1}}{{\sqrt a  - 3}}\\
 = \dfrac{{2\sqrt a  - 9 - \left( {\sqrt a  + 3} \right)\left( {\sqrt a  - 3} \right) + \left( {2\sqrt a  + 1} \right)\left( {\sqrt a  - 2} \right)}}{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  - 3} \right)}}\\
 = \dfrac{{2\sqrt a  - 9 - a + 9 + 2a - 3\sqrt a  - 2}}{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  - 3} \right)}}\\
 = \dfrac{{a - \sqrt a  - 2}}{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  - 3} \right)}}\\
 = \dfrac{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  + 1} \right)}}{{\left( {\sqrt a  - 2} \right)\left( {\sqrt a  - 3} \right)}}\\
 = \dfrac{{\sqrt a  + 1}}{{\sqrt a  - 3}}\\
2)A < 1\\
 \Rightarrow \dfrac{{\sqrt a  + 1}}{{\sqrt a  - 3}} < 1\\
 \Rightarrow \dfrac{{\sqrt a  + 1 - \sqrt a  + 3}}{{\sqrt a  - 3}} < 0\\
 \Rightarrow \dfrac{4}{{\sqrt a  - 3}} < 0\\
 \Rightarrow \sqrt a  - 3 < 0\\
 \Rightarrow \sqrt a  < 3\\
 \Rightarrow a < 9\\
Vay\,0 \le a < 9;a \ne 4\\
3)A = \dfrac{{\sqrt a  + 1}}{{\sqrt a  - 3}} = \dfrac{{\sqrt a  - 3 + 4}}{{\sqrt a  - 3}}\\
 = 1 + \dfrac{4}{{\sqrt a  - 3}} \in Z\\
 \Rightarrow \dfrac{4}{{\sqrt a  - 3}} \in Z\\
 \Rightarrow \left( {\sqrt a  - 3} \right) \in \left\{ { - 2; - 1;1;2;4} \right\}\\
 \Rightarrow \sqrt a  \in \left\{ {1;2;4;5;7} \right\}\\
 \Rightarrow a \in \left\{ {1;4;16;25;49} \right\}\\
Do:a \ne 4\\
 \Rightarrow a \in \left\{ {1;16;25;49} \right\}\\
B2:\\
1)\left( d \right)//y =  - x + 6\\
 \Rightarrow \left\{ \begin{array}{l}
a =  - 1\\
b \ne 6
\end{array} \right.\\
 \Rightarrow \left( d \right):y =  - x + b\\
A\left( { - 1; - 9} \right) \in \left( d \right)\\
 \Rightarrow  - 9 = 1 + b\\
 \Rightarrow b =  - 10\left( {tmdk} \right)\\
Vay\,a =  - 1;b =  - 10\\
2)\left\{ \begin{array}{l}
3x + 4y = 7\\
2x - y = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
3x + 4y = 7\\
8x - 4y = 4
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
11x = 11\\
y = 2x - 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
x = 1\\
y = 1
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left( {1;1} \right)
\end{array}$