$\color{blue}{P(A\hspace{0.1cm} \cap\hspace{0.1cm} B) = 0.30}$
$\color{blue}{P(A\hspace{0.1cm} \cap \hspace{0.1cm} C) = 0.20}$
Now, $B$ and $C$ are collectively exhaustive means $B$ & $C$ together exhaust all the possibilities.
∴ The union of $B$ & $C$ must cover all the events in the entire sample space.
∴ $\color{violet}{P(B \cup C) = 1}$
$B $ & $C$ are also mutually exclusive means $B $ & $C$ cannot occur simultaneously.
∴ $\color{violet}{P(B \cap C) = 0}$
${P(A\hspace{0.1cm} and\hspace{0.1cm} B) \neq P(A) \times P(B)}\Rightarrow$ because Event $B$ depends on Event $A$.
& ${P(A\hspace{0.1cm} and\hspace{0.1cm} C) \neq P(A) \times P(C)}\Rightarrow$ because Event $C$ depends on Event $A$.
Events A and B are independent events if and only if :
$P(A \cap B) = P(A) \times P(B)$
Otherwise, A and B are called dependent events.
Applying union in both the cases:
$P((A \cap B) \cup (A \cap C)) = 0.3 + 0.2 = 0.5$
As per distributive law
$P(A \cap (B \cup C)) = 0.5$
We already know that, $P(B \cup C) = 1$ as they are collectively exhaustive.
$∴P(A \cap 1) $ or $\color{violet}{P(A) = 0.5} $
$\color{green}{P(C/A) = \dfrac{P(C\cap A)}{P(A)} = \dfrac{0.20}{0.50}} =\color{orange} {0.4}$