$\begin{array}{l}
\cot a = - 3 \Rightarrow \tan a = \dfrac{1}{{\cot a}} = \dfrac{{ - 1}}{3}\\
\dfrac{{2\sin a - 3\cos a}}{{4\sin a + 5\cos a}} = \dfrac{{2\tan a.\cos a - 3\cos a}}{{4\tan a.\cos a + 5\cos a}} = \dfrac{{2\tan a - 3}}{{4\tan a + 5}} = \dfrac{{\dfrac{{ - 2}}{3} - 3}}{{\dfrac{{ - 4}}{3} + 5}} \\= - 1
\end{array}$