For events $A$, $B$ and $C$, it is given that $P(A)=0.7$, $P(B)=0.5$, $P(C\mid A')=0.6$ and $P(A\mid C')=0.76$.

1. Find the greatest possible value of $P(A \cap B)$.

The probability of the intersection of two events cannot be greater than the probability of either individual event.
$P(A \cap B) \le P(A)$
$P(A \cap B) \le P(B)$
$P(A \cap B) \le \min(0.7, 0.5) = 0.5$
Therefore, the greatest possible value is $0.5$.

2. Find the least possible value of $P(A \cap B)$.

The probability of the union of two events cannot exceed $1$.
Using the addition rule, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$P(A \cup B) \le 1$
$0.7 + 0.5 - P(A \cap B) \le 1$
$1.2 - P(A \cap B) \le 1$
$P(A \cap B) \ge 0.2$
Therefore, the least possible value is $0.2$.

1. Find the probability of the complement of $A$

$P(A') = 1 - P(A) = 1 - 0.7 = 0.3$

2. Use the conditional probability definition to find $P(C \cap A')$

$P(C \cap A') = P(C \mid A') \times P(A')$
$P(C \cap A') = 0.6 \times 0.3 = 0.18$

1. Find the probabilities of the complementary events $A'$ and $C' \mid A'$.

$P(A') = 1 - P(A) = 1 - 0.7 = 0.3$
$P(C' \mid A') = 1 - P(C \mid A') = 1 - 0.6 = 0.4$

2. Calculate $P(C' \cap A')$ using the multiplication rule for conditional probability.

$P(C' \cap A') = P(C' \mid A') \times P(A')$
$P(C' \cap A') = 0.4 \times 0.3 = 0.12$

1. Find $P(A' \cap C')$ (building upon earlier parts)

$P(A') = 1 - P(A) = 1 - 0.7 = 0.3$

$P(A' \cap C') = P(C' \mid A') P(A') = (1 - 0.6) \times 0.3 = 0.12$

2. Set up an equation for $P(C')$ using the definition of conditional probability

Since $C'$ is partitioned into mutually exclusive events $(A \cap C')$ and $(A' \cap C')$, we have:
$P(A \cap C') = P(C') - P(A' \cap C') = P(C') - 0.12$

Using $P(A \mid C') = 0.76$:
$\frac{P(A \cap C')}{P(C')} = 0.76$
$\frac{P(C') - 0.12}{P(C')} = 0.76$

3. Solve for $P(C')$ and hence find $P(C)$

$1 - \frac{0.12}{P(C')} = 0.76$

$\frac{0.12}{P(C')} = 0.24$

$P(C') = \frac{0.12}{0.24} = 0.5$

Therefore, $P(C) = 1 - P(C') = 1 - 0.5 = 0.5$