For events $A$ and $B$, it is given that $\mathrm{P}(A) = 0.38$, $\mathrm{P}(A \mid B) = 0.5$ and $\mathrm{P}(A \cup B) = 0.52$.

1. Express the probability of the intersection, $\mathrm{P}(A \cap B)$, in terms of $\mathrm{P}(B)$.

Using the conditional probability formula $\mathrm{P}(A \cap B) = \mathrm{P}(A \mid B)\mathrm{P}(B)$:
$\mathrm{P}(A \cap B) = 0.5\mathrm{P}(B)$

2. Set up an equation for $\mathrm{P}(B)$ using the addition rule.

Using the addition rule $\mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A \cap B)$:
$0.52 = 0.38 + \mathrm{P}(B) - 0.5\mathrm{P}(B)$
$0.52 = 0.38 + 0.5\mathrm{P}(B)$

3. Solve the linear equation for $\mathrm{P}(B)$.

$0.5\mathrm{P}(B) = 0.52 - 0.38$
$0.5\mathrm{P}(B) = 0.14$
$\mathrm{P}(B) = \frac{0.14}{0.5} = 0.28$

1. Apply De Morgan's laws to simplify the event given in the conditional probability.

$A' \cup B' = (A \cap B)'$
$\mathrm{P}((A \cap B)' \mid C) = 0.85$

2. Use the complement rule for conditional probabilities to find $\mathrm{P}(A \cap B \mid C)$.

$\mathrm{P}(A \cap B \mid C) = 1 - \mathrm{P}((A \cap B)' \mid C)$
$\mathrm{P}(A \cap B \mid C) = 1 - 0.85 = 0.15$

3. Apply the conditional probability formula to find the intersection of all three events.

$\mathrm{P}(A \cap B \cap C) = \mathrm{P}(A \cap B \mid C) \times \mathrm{P}(C)$
$\mathrm{P}(A \cap B \cap C) = 0.15 \times 0.6 = 0.09$

1. Establish the condition for mutually exclusive events and its implication on the three-event intersection.

If events $A$ and $C$ are mutually exclusive, then $\mathrm{P}(A \cap C) = 0$.
Since $(A \cap B \cap C)$ is a subset of $(A \cap C)$, it must follow that $\mathrm{P}(A \cap B \cap C) = 0$ as well.

2. Compare with the result from part (ii) to make a conclusion.

From part (ii), we found that $\mathrm{P}(A \cap B \cap C) = 0.09$.
Since $0.09 \neq 0$, this implies $\mathrm{P}(A \cap C) \ge 0.09$, and thus $\mathrm{P}(A \cap C) \neq 0$.
Therefore, events $A$ and $C$ are not mutually exclusive.