A laboratory tests water samples for a pollutant…

Step 1. The condition. $P(\text{at least one}) = 1 - 0.92^n$. For $95\%$ we need $0.92^n \le 0.05$.

Step 2. Solve. $n \ge \dfrac{\ln 0.05}{\ln 0.92} \approx 35.93$. Round up: $n = 36$.

Step 3. For $99\%$. $n \ge \dfrac{\ln 0.01}{\ln 0.92} \approx 55.23$. Round up: $n = 56$.

Step 4. The probability for $36$ samples. $1 - 0.92^{36} \approx 0.9503$. With $35$ samples it would be $0.9460$, which is below $0.95$.