All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\mathrm{P}(A|B') = \frac{\mathrm{P} (A \cap B')}{\mathrm{P}(B')}$
$\Rightarrow \mathrm{P} (A \cap B') = 0.8 \times (1-0.6) = 0.32$

(ii)
$\mathrm{P}(A \cup B) = \mathrm{P}(A \cap B') + \mathrm{P}(B) = 0.32 + 0.6 = 0.92$

(iii)
P(B’|A) $= \frac{\mathrm{P} (B' \cap A)}{\mathrm{P}(A)} = \frac{0.32}{0.7}= 0.457$

(iv)
$\mathrm{P} (A' \cap C) = \mathrm{P} (A')\mathrm{P} (C)$ since A and C are independent
$\Rightarrow \mathrm{P} (A' \cap C) = 0.15$

(v)
$0 \le \mathrm{P} (A' \cap B \cap C) \le 0.15$