事件 $A$ 和 $B$ 满足 $\text{P}(A) = a$ 且 $\text{P}(B) = b$。 $A$ 和 $B$ 是互斥事件。

1. Relate $\text{P}(A' \cap B')$ to the union of events $A$ and $B$.

$\text{P}(A' \cap B') = 1 - \text{P}(A \cup B)$

2. Apply the mutually exclusive property to substitute probabilities in terms of $a$ and $b$.

Since $A$ and $B$ are mutually exclusive events, $\text{P}(A \cap B) = 0$.
$\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) = a + b$
Therefore,
$\text{P}(A' \cap B') = 1 - (a + b)$

1. Describe the required Venn diagram for the given conditions

Since $A$ and $B$ are mutually exclusive, their regions do not overlap. Since $A'$ and $B'$ are mutually exclusive, $A' \cap B' = \emptyset$, meaning the region outside $A$ and $B$ is empty.
The Venn diagram should be a rectangular universal set box, divided entirely into two disjoint sections by a line, labelling one section $A$ and the other $B$. (Alternatively, two disjoint circles for $A$ and $B$ can be drawn inside the rectangle, with the region outside the circles explicitly labelled as $0$ or $\emptyset$).

2. State the equation relating $a$ and $b$

Since $A'$ and $B'$ are mutually exclusive, $\text{P}(A' \cap B') = 0$.
From the properties of mutually exclusive sets $A$ and $B$, $\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) = a + b$.
Since $\text{P}(A' \cap B') = 1 - \text{P}(A \cup B)$:
$1 - (a + b) = 0$
$a + b = 1$

1. Determine the probabilities of intersection and union for the basic events using independence and mutual exclusivity.

Since $A$ and $C$ are independent:
$\text{P}(A \cap C) = \text{P}(A)\text{P}(C) = 0.2c$

Since $A$ and $B$ are mutually exclusive:
$\text{P}(A \cup B) = \text{P}(A) + \text{P}(B) = 0.2 + 0.3 = 0.5$

2. Apply the addition rule to find $\text{P}(A \cup B \cup C)$.

Since $A$ and $B$ are mutually exclusive, $(A \cap C)$ and $(B \cap C)$ are also mutually exclusive. Hence:
$\text{P}((A \cup B) \cap C) = \text{P}(A \cap C) + \text{P}(B \cap C) = 0.2c + 0.14$

Using the addition rule for the events $(A \cup B)$ and $C$:
$\text{P}(A \cup B \cup C) = \text{P}(A \cup B) + \text{P}(C) - \text{P}((A \cup B) \cap C)$
$\text{P}(A \cup B \cup C) = 0.5 + c - (0.2c + 0.14)$
$\text{P}(A \cup B \cup C) = 0.36 + 0.8c$

1. Express the required probability by subtracting the overlaps with $A$ and $B$ from the total probability of $C$.

The event $A' \cap B' \cap C$ represents the outcome where $C$ occurs, but neither $A$ nor $B$ occurs. Since $A$ and $B$ are mutually exclusive, their intersections with $C$ do not overlap.
$\text{P}(A' \cap B' \cap C) = \text{P}(C) - \text{P}(A \cap C) - \text{P}(B \cap C)$

2. Use the independence of $A$ and $C$ and substitute the given values to simplify the expression.

Since $A$ and $C$ are independent events, $\text{P}(A \cap C) = \text{P}(A)\text{P}(C) = 0.2c$.
We are given $\text{P}(C) = c$ and $\text{P}(B \cap C) = 0.14$.
$\text{P}(A' \cap B' \cap C) = c - 0.2c - 0.14$
$= 0.8c - 0.14$

1. Concisely obtain the expression for part (iii) and use the maximum probability bound to find an upper limit for $c$.

First, we determine the expression for $\text{P}(A \cup B \cup C)$ as required in (iii). Using the inclusion-exclusion principle and the independence of $A$ and $C$ (so $\text{P}(A \cap C) = 0.2c$):
$\text{P}(A \cup B \cup C) = \text{P}(A) + \text{P}(B) + \text{P}(C) - \text{P}(A \cap B) - \text{P}(B \cap C) - \text{P}(A \cap C) + \text{P}(A \cap B \cap C)$
Since $A$ and $B$ are mutually exclusive, $\text{P}(A \cap B) = 0$ and consequently $\text{P}(A \cap B \cap C) = 0$.
$\text{P}(A \cup B \cup C) = 0.2 + 0.3 + c - 0 - 0.14 - 0.2c + 0 = 0.36 + 0.8c$
Since the maximum possible value for any probability is $1$, we have:
$0.36 + 0.8c \le 1$
$0.8c \le 0.64 \implies c \le 0.8$

2. Concisely obtain the expression for part (iv) and use the minimum probability bound to find a lower limit for $c$.

Next, we determine the expression for $\text{P}(A' \cap B' \cap C)$ as required in (iv). This event represents occurring in $C$ but not in $A$ nor $B$:
$\text{P}(A' \cap B' \cap C) = \text{P}(C) - \text{P}(C \cap (A \cup B))$
Since $A$ and $B$ are mutually exclusive, $\text{P}(C \cap (A \cup B)) = \text{P}(A \cap C) + \text{P}(B \cap C)$.
$\text{P}(A' \cap B' \cap C) = c - (0.2c + 0.14) = 0.8c - 0.14$
Since probabilities cannot be negative, we have:
$0.8c - 0.14 \ge 0$
$0.8c \ge 0.14 \implies c \ge 0.175$

3. Combine the bounds to state the full range of values for $c$.

Combining the inequalities gives the complete range:
$0.175 \le c \le 0.8$