A number is chosen at random from the whole numbers 1 to 60…

Step 1. The counts. $n(A) = 20$, $n(B) = 15$ and $n(C) = 12$. The pairs: $n(A \cap B) = 5$ (multiples of $12$), $n(B \cap C) = 3$ (multiples of $20$) and $n(A \cap C) = 4$ (multiples of $15$). The triple: $n(A \cap B \cap C) = 1$ (the multiples of $60$).

Step 2. At least one. $20 + 15 + 12 - 5 - 3 - 4 + 1 = 36$, so $P = \dfrac{36}{60} = 0.6$.

Step 3. None. $1 - 0.6 = 0.4$, which is $24$ numbers.

Step 4. Exactly one. Only $A$: $20 - 5 - 4 + 1 = 12$. Only $B$: $15 - 5 - 3 + 1 = 8$. Only $C$: $12 - 3 - 4 + 1 = 6$. The total is $26$, so $P = \dfrac{26}{60} \approx 0.433$.

Step 5. All three. $\dfrac{1}{60} \approx 0.017$.