Divide the sum of $\dfrac{13}{5} \text{ and } \dfrac{-12}{7}$ by the product of $\dfrac{-31}{7} \text{ and } \dfrac{1}{-2}$.

First find the sum of $\dfrac{13}{5} \text{ and } \dfrac{-12}{7}$ and the product of $\dfrac{-31}{7} \text{ and } \dfrac{1}{-2}$.

Note that LCM of 5 and 7 is 35.

Finding the sum,

$\dfrac{13}{5} + \dfrac{-12}{7} = \dfrac{13 \times 7 + (-12) \times 5}{35} \\[1em] = \dfrac{91 - 60}{35} \\[1em] = \dfrac{31}{35}$

Finding the product,

$\dfrac{-31}{7} \times \dfrac{1}{-2} = \dfrac{(-31) \times 1}{7 \times (-2)} \\[1em] = \dfrac{-31}{-14} = \dfrac{31}{14}$

Dividing the sum by the product,

$\dfrac{31}{35} \div \dfrac{31}{14} = \dfrac{31}{35} \times \dfrac{14}{31} \\[1em] = \dfrac{31 \times 14}{35 \times 31} \\[1em] = \dfrac{14}{35} = \dfrac{2}{5}$

Hence, the required quotient is $\mathbf{\dfrac{2}{5}}$.