The events $A$ and $B$ are such that $\mathrm{P}(B) = \frac{1}{3}$, $\mathrm{P}(A|B) = \frac{2}{5}$ and $\mathrm{P}(A \cup B) = \frac{3}{5}$.

1. Calculate the probability of the intersection of A and B using the conditional probability formula.

$\mathrm{P}(A \cap B) = \mathrm{P}(A|B) \times \mathrm{P}(B) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$

2. Find $\mathrm{P}(A)$ using the addition rule of probability.

$\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$
$\frac{3}{5} = \mathrm{P}(A) + \frac{1}{3} - \frac{2}{15}$
$\mathrm{P}(A) = \frac{3}{5} - \frac{1}{3} + \frac{2}{15} = \frac{9 - 5 + 2}{15} = \frac{6}{15} = \frac{2}{5}$

3. Determine if events A and B are independent by checking the mathematical conditions for independence.

Since $\mathrm{P}(A|B) = \frac{2}{5}$ and $\mathrm{P}(A) = \frac{2}{5}$, we observe that $\mathrm{P}(A|B) = \mathrm{P}(A)$.

Alternatively, checking the product rule:
$\mathrm{P}(A) \times \mathrm{P}(B) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$
Since $\mathrm{P}(A) \times \mathrm{P}(B) = \mathrm{P}(A \cap B)$, events $A$ and $B$ are independent.

1. Calculate the probability of the intersection of $A$ and $B$.

$\mathrm{P}(A \cap B) = \mathrm{P}(A|B)\mathrm{P}(B) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$

2. Calculate $\mathrm{P}(A' \cap B)$ using the rule for the difference of events.

$\mathrm{P}(A' \cap B) = \mathrm{P}(B) - \mathrm{P}(A \cap B) = \frac{1}{3} - \frac{2}{15} = \frac{5}{15} - \frac{2}{15} = \frac{3}{15} = \frac{1}{5}$

1. Define variables for the probabilities of the relevant regions within $A'$ to formulate equations.

Since $A$ and $C$ are mutually exclusive, $\mathrm{P}(A \cap C) = 0$. This means event $C$ occurs entirely within $A'$.
Let $y = \mathrm{P}(A' \cap B' \cap C)$. Since $\mathrm{P}(A \cap B' \cap C) = 0$, $y = \mathrm{P}(B' \cap C)$.
Let $x = \mathrm{P}(A' \cap B \cap C)$.
Event $C$ is partitioned into its intersections with $B$ and $B'$, so:
$x + y = \mathrm{P}(C) = \frac{1}{2}$

2. Use the probability of $A' \cap B$ and the non-negativity of probabilities to bound $x$ and find the least value of $y$.

From part (ii), we have $\mathrm{P}(A' \cap B) = \frac{1}{5}$.
The region $A' \cap B$ is partitioned by $C$ and $C'$. Thus, the probability of the part outside $C$ is:
$\mathrm{P}(A' \cap B \cap C') = \frac{1}{5} - x$
Since probabilities cannot be negative, $\frac{1}{5} - x \ge 0$, which gives $x \le \frac{1}{5}$.
Substitute $x = \frac{1}{2} - y$:
$\frac{1}{2} - y \le \frac{1}{5} \implies y \ge \frac{1}{2} - \frac{1}{5} = \frac{3}{10}$
The least possible value of $\mathrm{P}(B' \cap C)$ is $\frac{3}{10}$.

3. Use the total probability of $A'$ to bound $y$ and find its greatest value.

From part (i), we established $\mathrm{P}(A) = \frac{2}{5}$, so $\mathrm{P}(A') = 1 - \frac{2}{5} = \frac{3}{5}$.
Let $z = \mathrm{P}(A' \cap B' \cap C')$.
The event $A'$ comprises four mutually exclusive regions, so their probabilities must sum to $\mathrm{P}(A')$:
$\mathrm{P}(A' \cap B \cap C) + \mathrm{P}(A' \cap B \cap C') + \mathrm{P}(A' \cap B' \cap C) + \mathrm{P}(A' \cap B' \cap C') = \mathrm{P}(A')$
$x + \left(\frac{1}{5} - x\right) + y + z = \frac{3}{5}$
$\frac{1}{5} + y + z = \frac{3}{5} \implies y + z = \frac{2}{5}$
Since $z \ge 0$, we must have $y \le \frac{2}{5}$.
The greatest possible value of $\mathrm{P}(B' \cap C)$ is $\frac{2}{5}$.