I had attempted a solution to this question, but I didn't get anywhere fruitful. I did find a solution and I'm going to copy it here because there are some steps I'm not clear behind why they make sense and wanted clarification on them.
i) Proving $P((A \cap B) \cap C) = P(A \cap B) \cap P(C) = P(A)P(B)P(C)$ (since the elements are mutually independent.)
Pf: Consider $$P((A \cap B)|C) = \frac{P((A \cap B) \cap C)}{P(C)} \\ = \frac{P(A)P(B)P(C)}{P(C)} \\ = P(A)P(B) \\ = P(A \cap B)$$
Hence $A \cap B)$ and $C$ are independent.
Question: Why is the person who wrote the solution conditioning on $C$ and how does this whole conditioning even imply independence?
ii) Proving $P((A \cup B) \cap C) = P(A \cup B) P(C)$
Pf: Consider $$P((A \cup B)|C) = \frac{P((A \cup B) \cap C)}{P(C)} \\ \frac{P((A \cap C) \cup (B \cap C))}{P(C)} \\ = \frac{P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)}{P(C)} \\ = \frac{P(A)P(C) + P(B)P(C) - P(A)P(B)P(C)}{P(C)} \\ = \frac{P(C)[P(A) + P(B) - P(A)P(B)]}{P(C)} \\ = P(A) + P(B) - P(A)P(B) \\ = P(A \cup B)$$
Henc $A \cup B$ and $C$ are independent.
Question: Again how does complementing accomplish our intended result and why does it work?
Final Observation: After writing out the solutions I did notice that between the start of the chain of equalities and the ending one, I could "multiply through" by $P(C)$ to get my final result of independence, thus for example in i we started with and ended with
$$\frac{P((A \cap B) \cap C)}{P(C)} = P(A \cap B)$$
So from here I could "multiply through" and would end up with
$$ P(A \cap B)P(C)$$
Mechanically I suppose it is correct, but it doesn't sit well with me if that is how to get the end result. Particularly I'm still troubled by the idea of conditioning. Could somebody add clarification to my issue?