The events $A$, $B$ and $C$ are such that $P(A)=0.8$, $P(B)=0.2$ and $P(C)=0.6$. It is also known that $P(A\cap B)=P(B)$ and $P(B\cap C)=0.1$.

1. Determine $P(A \cap B \cap C)$ using the given subset condition.

Since $P(A \cap B) = P(B) = 0.2$, the event $B$ must be a subset of $A$ ($B \subseteq A$).
Therefore, the intersection of $B$ and $C$ is entirely contained within $A$, meaning $A \cap B \cap C = B \cap C$.
$P(A \cap B \cap C) = P(B \cap C) = 0.1$

2. Express $P(A \cup C)$ in terms of $P(A \cap B^c \cap C)$.

We partition $A \cap C$ into two mutually exclusive events based on $B$:
$P(A \cap C) = P(A \cap B \cap C) + P(A \cap B^c \cap C)$
$P(A \cap C) = 0.1 + P(A \cap B^c \cap C)$
Using the addition rule for probabilities:
$P(A \cup C) = P(A) + P(C) - P(A \cap C)$
$P(A \cup C) = 0.8 + 0.6 - (0.1 + P(A \cap B^c \cap C))$
$P(A \cup C) = 1.3 - P(A \cap B^c \cap C)$

3. Apply probability bounds to find the maximum and minimum values.

Since any probability must not exceed $1$, we have $P(A \cup C) \le 1$:
$1.3 - P(A \cap B^c \cap C) \le 1$
$P(A \cap B^c \cap C) \ge 0.3$
Also, since $A \cap C \subseteq C$, its probability is bounded by $P(C)$:
$P(A \cap C) \le P(C)$
$0.1 + P(A \cap B^c \cap C) \le 0.6$
$P(A \cap B^c \cap C) \le 0.5$
Thus, the maximum possible value is $0.5$ and the minimum possible value is $0.3$.

1. Determine the required range of values for $P(A \cap B^c \cap C)$ from the previously given probabilities.

Let $x = P(A \cap B^c \cap C)$.
Since $P(A \cap B) = P(B) = 0.2$, the event $B$ is a subset of $A$, meaning $A \cap B \cap C = B \cap C$.
Therefore, $P(A \cap C) = P(A \cap B \cap C) + P(A \cap B^c \cap C) = 0.1 + x$.
Using the addition rule, $P(A \cup C) = P(A) + P(C) - P(A \cap C) = 0.8 + 0.6 - (0.1 + x) = 1.3 - x$.
Since any probability cannot exceed $1$, $P(A \cup C) \le 1 \implies 1.3 - x \le 1 \implies x \ge 0.3$.
Additionally, since $A \cap B^c \cap C \subseteq B^c \cap C$, we have $x \le P(C) - P(B \cap C) = 0.6 - 0.1 = 0.5$.
Thus, $0.3 \le x \le 0.5$.

2. Express the required conditional probability in terms of $x$.

By the definition of conditional probability,
$P(A \cap B^c \cap C \mid A \cup C) = \frac{P(A \cap B^c \cap C \cap (A \cup C))}{P(A \cup C)}$
Since $A \cap B^c \cap C$ is a subset of $A \cup C$, the intersection is simply $A \cap B^c \cap C$, making the numerator $P(A \cap B^c \cap C) = x$.
From Step 1, we established that $P(A \cup C) = 1.3 - x$.
Thus, the conditional probability expression simplifies to $\frac{x}{1.3 - x}$.

3. Find the maximum and minimum of the conditional probability using the feasible range of $x$.

We define the function $\mathrm{f}(x) = \frac{x}{1.3 - x}$ over the domain $0.3 \le x \le 0.5$.
We can rewrite this algebraically:
$\frac{x}{1.3 - x} = \frac{-(1.3 - x) + 1.3}{1.3 - x} = -1 + \frac{1.3}{1.3 - x}$
As $x$ increases, the denominator $(1.3 - x)$ decreases, causing the fraction $\frac{1.3}{1.3 - x}$ to increase. Hence, the function is strictly increasing over the interval $[0.3, 0.5]$.
The minimum possible value occurs at $x = 0.3$:
$\text{Minimum} = \frac{0.3}{1.3 - 0.3} = \frac{0.3}{1} = \frac{3}{10}$
The maximum possible value occurs at $x = 0.5$:
$\text{Maximum} = \frac{0.5}{1.3 - 0.5} = \frac{0.5}{0.8} = \frac{5}{8}$

1. Calculate the probability of the intersection of independent events $A$ and $C$.

Since $A$ and $C$ are independent,
$P(A \cap C) = P(A)P(C) = 0.8 \times 0.6 = 0.48$

2. Express $P(A \cap C)$ in terms of its disjoint subsets and solve for the required probability.

Given $P(A \cap B) = P(B) = 0.2$, we have $P(A' \cap B) = 0$.
This implies that $P(A' \cap B \cap C) = 0$.
Since $P(B \cap C) = P(A \cap B \cap C) + P(A' \cap B \cap C)$,
$P(A \cap B \cap C) = 0.1 - 0 = 0.1$
Now, decomposing $P(A \cap C)$ into disjoint subsets involving $B$ and $B^c$:
$P(A \cap C) = P(A \cap B \cap C) + P(A \cap B^c \cap C)$
$0.48 = 0.1 + P(A \cap B^c \cap C)$
$P(A \cap B^c \cap C) = 0.48 - 0.1 = 0.38$