### Problem:

A small town's water treatment facility monitors the presence of a particular toxin in its water supply. The toxin appears in the water randomly and independently over time. Historical data indicate that the toxin appears at an average rate of 3 times every 2 days. The town's water treatment manager wants to calculate the probability that the toxin will appear:

1. Exactly 5 times in a 2-day period.
2. At least 2 times but no more than 4 times in a 2-day period.

The appearance of the toxin can be modeled as a Poisson process, a common approach for counting the number of events (in this case, toxin appearances) that occur within a fixed interval of time, area, or volume.

### Solution:

1. **Modeling the Problem:**

We can model the appearance of the toxin using a Poisson distribution. If $\lambda$ represents the average number of appearances in a 2-day period, the probability of observing exactly $k$ appearances is given by:

$$
P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}
$$

Given that the average rate $\lambda$ is 3 appearances every 2 days.

2. **Calculating the Probability of Exactly 5 Appearances:**

We need to find $P(X = 5)$ with $\lambda = 3$:

$$
P(X = 5) = \frac{e^{-3} \cdot 3^5}{5!}
$$

Let's compute this:

$$
P(X = 5) = \frac{e^{-3} \cdot 243}{120} \approx 0.1008
$$

So, the probability of exactly 5 appearances in a 2-day period is approximately $10.08\%$.

3. **Calculating the Probability of At Least 2 but No More Than 4 Appearances:**

We need to find $P(2 \leq X \leq 4)$. This is equal to $P(X = 2) + P(X = 3) + P(X = 4)$:

$$
P(2 \leq X \leq 4) = \frac{e^{-3} \cdot 3^2}{2!} + \frac{e^{-3} \cdot 3^3}{3!} + \frac{e^{-3} \cdot 3^4}{4!}
$$

Let's compute this:

$$
P(2 \leq X \leq 4) = e^{-3} \left( \frac{9}{2} + \frac{27}{6} + \frac{81}{24} \right)
$$

$$
P(2 \leq X \leq 4) \approx 0.61
$$

So, the probability of observing the toxin at least 2 times but no more than 4 times in a 2-day period is approximately $22.40\%$.