Let $P$ and $Q$ be the points of trisection of $A B$.

=> P divides $A B$ in the radio $1: 2$

So, the coordinates of $P$ are

$$
\begin{aligned}
&x=\frac{\left(m x_{2}+n x_{1}\right)}{(m+n)}, y=\frac{\left(m y_{2}+n y_{1}\right)}{(m+n)} \\
&\Rightarrow x=\frac{\{1 \times 1+2 \times(3)\}}{1+2}, y=\frac{\{1 \times 2+2 \times(-4)\}}{1+2} \\
&\Rightarrow x=\frac{1+6}{3}, y=\frac{2-8}{3} \\
&\Rightarrow x=\frac{7}{3}, y=-\frac{6}{3} \\
&\Rightarrow x=\frac{7}{3}, y=-2
\end{aligned}
$$

=> the coordinates of $\mathrm{P}$ are $\left(\frac{7}{3},-2\right)$

$(p,-2)$ are the coordinates of $P$.

$p=\frac{7}{3}$

Q divides the line AB in the ratio $2: 1$

=> the coordinates of $Q$ are

$$
\begin{aligned}
&\mathrm{x}=\frac{\left(\mathrm{mx}_{2}+\mathrm{mx}_{1}\right)}{(\mathrm{m}+\mathrm{n})}, \mathrm{y}=\frac{\left(\mathrm{my}_{2}+\mathrm{my}_{1}\right)}{(\mathrm{m}+\mathrm{n})} \\
&\Rightarrow \mathrm{x}=\frac{(2 \times 1+1 \times 3)}{2+1}, \mathrm{y}=\frac{\{2 \times 2+1 \times(-4)\}}{2+1} \\
&\Rightarrow \mathrm{x}=\frac{2+3}{3}, \mathrm{y}=\frac{4-4}{3} \\
&\Rightarrow \mathrm{x}=\frac{5}{3}, \mathrm{y}=0
\end{aligned}
$$

=> coordinates of $Q$ are $\left(\frac{5}{3}, 0\right)$.

But the given coordinates of $Q$ are $\left(\frac{5}{3}, q\right)$

$\mathrm{So}, \mathrm{q}=0$

Therefore, $p=\frac{7}{3}$ and $q=0$