1.

a) $\frac{{ - 2}}{3} + \frac{4}{5}\, + \frac{{ - 5}}{3} - \frac{{19}}{5} = \left( {\frac{{ - 2}}{3} + \frac{{ - 5}}{3}} \right) + \left( {\frac{4}{5}\, - \frac{{19}}{5}} \right)$

$= \frac{{ - 7}}{3} + \left( { - 3} \right) = \frac{{ - 7}}{3} - \frac{9}{3} = \frac{{ - 16}}{3}$.                                   

b) $12\,\,.\,\,{\left( {\frac{3}{2}} \right)^2} + {\left( {\frac{{25}}{6}} \right)^0}\,\,.\,\,\frac{{15}}{2} - 9\frac{1}{4} = 12\,\,.\,\,\frac{9}{4} + 1\,\,.\,\,\frac{{15}}{2} - \frac{{37}}{4}$

$= 12\,\,.\,\,\frac{9}{4} + 1\,\,.\,\,\frac{{15}}{2} - \frac{{37}}{4} = 27 + \frac{{15}}{2} - \frac{{37}}{4} = \frac{{69}}{2} - \frac{{37}}{4} = \frac{{101}}{4}$.

2.

a) $\frac{{ - 3}}{8} + \frac{5}{6}x = \frac{{ - 17}}{{12}}$                                           

$\frac{5}{6}x = \frac{{ - 17}}{{12}} + \frac{3}{8}$

$\frac{5}{6}x = \frac{{ - 25}}{{24}}$

$x = \frac{{ - 25}}{{24}}:\frac{5}{6}$

$x = \frac{{ - 5}}{4}$.

Vậy $x = \frac{{ - 5}}{4}$.

b) ${\left( {\frac{1}{2}} \right)^3} - \left( {x + \frac{2}{5}} \right)\,:\frac{2}{3} = \frac{{ - 31}}{{40}}$

$\frac{1}{8} - \left( {x + \frac{2}{5}} \right)\,\,\,.\,\,\frac{3}{2} = \frac{{ - 31}}{{40}}$

$\left( {x + \frac{2}{5}} \right)\,\,\,.\,\,\frac{3}{2} = \frac{1}{8} + \frac{{31}}{{40}}$

$\left( {x + \frac{2}{5}} \right)\,\,\,.\,\,\frac{3}{2} = \frac{9}{{10}}$

$x + \frac{2}{5}\, = \frac{9}{{10}}:\,\frac{3}{2}$

$x + \frac{2}{5}\, = \frac{3}{5}$

$x\, = \frac{3}{5} - \frac{2}{5}$

$x\, = \frac{1}{5}$

Vậy $x\, = \frac{1}{5}$.