$$ \because $$ $$ \tan \theta = \dfrac{4}{3} $$ (given)
We know that
$$ 1 + \tan^{2} \theta = \sec^{2} \theta $$
$$ \sec \theta = \sqrt{1 + \tan^{2} \theta} $$
$$ = \sqrt{1 + \left ( \dfrac{4}{3} \right )^{2}} = \sqrt{1 + \dfrac{16}{9}} $$
$$ = \sqrt{\dfrac{25}{9}} = \dfrac{5}{3} $$
$$ \therefore $$ $$ \cos \theta = \dfrac{3}{5} $$
$$ \therefore $$ $$ \sin \theta = \cos \theta \times \tan \theta $$
$$ = \dfrac{3}{5} \times \dfrac{4}{5} = \dfrac{4}{5} $$
$$ \therefore $$ Value of the expression $$ = \dfrac{3 \sin \theta + 2 \cos \theta}{3 \sin \theta - 2 \cos \theta} $$
$$ = \dfrac{3 \times \dfrac{4}{5} + 2 \times \dfrac{3}{5}}{3 \times \dfrac{4}{5} - 2 \times \dfrac{3}{5}} = \dfrac{\dfrac{12}{5} + \dfrac{6}{5}}{\dfrac{12}{5} - \dfrac{6}{5}} $$
Hence, $$ \dfrac{3 \sin \theta + 2 \cos \theta}{3 \sin \theta - 2 \cos \theta} = \dfrac{18}{6} = 3 $$