Arrange in ascending order:

$-2, \dfrac{1}{3}, \dfrac{-13}{6} \text{ and } \dfrac{8}{-3}$

First write each rational number with a positive denominator, then write each of them with the LCM of the denominators as the common denominator and arrange the numerators in increasing order.

$\Rightarrow -2 = \dfrac{-2}{1} \\[1em] \Rightarrow \dfrac{8}{-3} = \dfrac{8 \times (-1)}{-3 \times (-1)} = \dfrac{-8}{3}$

So, the given rational numbers are $\dfrac{-2}{1}, \dfrac{1}{3}, \dfrac{-13}{6} \text{ and } \dfrac{-8}{3}$.

Now write each rational number with the LCM of the denominators as the common denominator.

LCM of 1, 3, 6 and 3 = 6

$\Rightarrow \dfrac{-2}{1} = \dfrac{(-2) \times 6}{1 \times 6} = \dfrac{-12}{6} \\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{1 \times 2}{3 \times 2} = \dfrac{2}{6} \\[1em] \Rightarrow \dfrac{-13}{6} = \dfrac{-13}{6} \\[1em] \Rightarrow \dfrac{-8}{3} = \dfrac{(-8) \times 2}{3 \times 2} = \dfrac{-16}{6}$

As -16 < -13 < -12 < 2,

so $\dfrac{-16}{6} \lt \dfrac{-13}{6} \lt \dfrac{-12}{6} \lt \dfrac{2}{6}$

$\Rightarrow \dfrac{-8}{3} \lt \dfrac{-13}{6} \lt -2 \lt \dfrac{1}{3}$

Hence, the given rational numbers in ascending order are $\mathbf{\dfrac{8}{-3}, \dfrac{-13}{6}, -2 \text{ and } \dfrac{1}{3}}$.