Đáp án:

$\begin{array}{l}
a)Dkxd:a \ne  - \dfrac{1}{3};a \ne  - 3\\
\dfrac{{3a - 1}}{{3a + 1}} - \dfrac{{a - 3}}{{a + 3}} = 2\\
 \Rightarrow \dfrac{{\left( {3a - 1} \right)\left( {a + 3} \right) - \left( {a - 3} \right)\left( {3a + 1} \right)}}{{\left( {3a + 1} \right)\left( {a + 3} \right)}} = 2\\
 \Rightarrow 3{a^2} + 9a - a - 3 - \left( {3{a^2} + a - 9a - 3} \right)\\
 = 2\left( {3{a^2} + 9a + a + 3} \right)\\
 \Rightarrow 18a - 2a = 6{a^2} + 20a + 6\\
 \Rightarrow 6{a^2} + 4a + 6 = 0\\
 \Rightarrow 3{a^2} + 2a + 3 = 0\\
 \Rightarrow {a^2} + 2.\dfrac{1}{3}.a + \dfrac{1}{9} + \dfrac{8}{9} = 0\\
 \Rightarrow {\left( {a + \dfrac{1}{3}} \right)^2} + \dfrac{8}{9} = 0\left( {vn} \right)
\end{array}$

Vậy ko có giá trị của a để biểu thức bằng 2

$\begin{array}{l}
b)Dkxd:a \ne \dfrac{5}{2};a \ne \dfrac{2}{3}\\
\dfrac{{2a - 9}}{{2a - 5}} + \dfrac{{3a}}{{3a - 2}} = 2\\
 \Rightarrow \dfrac{{2a - 5 - 4}}{{2a - 5}} + \dfrac{{3a - 2 + 2}}{{3a - 2}} = 2\\
 \Rightarrow 1 - \dfrac{4}{{2a - 5}} + 1 + \dfrac{2}{{3a - 2}} = 2\\
 \Rightarrow \dfrac{2}{{3a - 2}} - \dfrac{4}{{2a - 5}} = 0\\
 \Rightarrow \dfrac{2}{{3a - 2}} = \dfrac{4}{{2a - 5}}\\
 \Rightarrow 2a - 5 = 2\left( {3a - 2} \right)\\
 \Rightarrow 2a - 5 = 6a - 4\\
 \Rightarrow 6a - 2a =  - 5 + 4\\
 \Rightarrow 4a =  - 1\\
 \Rightarrow a = \dfrac{{ - 1}}{4}\left( {tmdk} \right)\\
Vậy\,a =  - \dfrac{1}{4}\\
c)Dkxd:a \ne  - 3\\
\dfrac{{10}}{3} - \dfrac{{3a - 1}}{{4a + 12}} - \dfrac{{7a + 2}}{{6a + 18}} = 2\\
 \Rightarrow \dfrac{{10}}{3} - \dfrac{{3a - 1}}{{4\left( {a + 3} \right)}} - \dfrac{{7a + 2}}{{6\left( {a + 3} \right)}} = 2\\
 \Rightarrow \dfrac{{10.4.\left( {a + 3} \right) - 3\left( {3a - 1} \right) - 2\left( {7a + 2} \right)}}{{12\left( {a + 3} \right)}} = 2\\
 \Rightarrow 40a + 120 - 9a + 3 - 14a - 4 = 24\left( {a + 3} \right)\\
 \Rightarrow 17a + 119 = 24a + 72\\
 \Rightarrow 7a = 47\\
 \Rightarrow a = \dfrac{{47}}{7}\left( {tmdk} \right)\\
Vậy\,a = \dfrac{{47}}{7}
\end{array}$