Step 1: Find the first trisection point $P$ using the section formula with ratio $1:2$.
For $A(-1, 4)$ and $B(-3, -2)$, point $P$ divides $AB$ so that $AP:PB = 1:2$.
\[ x_P = \frac{1(-3) + 2(-1)}{1+2} = \frac{-3-2}{3} = -\frac{5}{3} \]
\[ y_P = \frac{1(-2) + 2(4)}{1+2} = \frac{-2+8}{3} = 2 \]
So $P = \left(-\frac{5}{3}, 2\right)$.
Step 2: Get the second trisection point $Q$ as the midpoint of $P$ and $B$, instead of applying the section formula a second time.
Since $AP = PQ = QB$, the point $Q$ sits exactly halfway between $P$ and $B$. The midpoint formula gives:
\[ x_Q = \frac{x_P + x_B}{2} = \frac{-\frac{5}{3} + (-3)}{2} = \frac{-\frac{5}{3} - \frac{9}{3}}{2} = \frac{-\frac{14}{3}}{2} = -\frac{7}{3} \]
\[ y_Q = \frac{y_P + y_B}{2} = \frac{2 + (-2)}{2} = \frac{0}{2} = 0 \]
Step 3: State the coordinates of $Q$.
\[ Q = \left(-\frac{7}{3}, 0\right) \]
Step 4: Confirm the pattern is consistent.
$P$ is one third of the way from $A$ to $B$, and $Q$, being the midpoint of $P$ and $B$, automatically lands at two thirds of the way from $A$ to $B$, exactly what a trisection point should be.
Final Answer:
The points of trisection are $\left(-\frac{5}{3}, 2\right)$ and $\left(-\frac{7}{3}, 0\right)$, which corresponds to option (A).
\[ \boxed{\left(-\frac{5}{3}, 2\right) \text{ and } \left(-\frac{7}{3}, 0\right)} \]