(i) The probability of not getting a 3 on a single die is \(\frac{5}{6}\). Therefore, the probability of not getting a 3 on all 9 dice is \(\left( \frac{5}{6} \right)^9\). The probability of getting at least one 3 is:

\(1 - \left( \frac{5}{6} \right)^9 = 1 - 0.1938 = 0.806\)

(ii) The probability of getting at least one 3 when \(n\) dice are thrown is \(1 - \left( \frac{5}{6} \right)^n\). We need this probability to be greater than 0.9:

\(1 - \left( \frac{5}{6} \right)^n > 0.9\)

\(\left( \frac{5}{6} \right)^n < 0.1\)

Taking logarithms on both sides:

\(n \log \left( \frac{5}{6} \right) < \log(0.1)\)

\(n > \frac{\log(0.1)}{\log \left( \frac{5}{6} \right)}\)

Calculating gives \(n > 12.6\), so the smallest integer \(n\) is 13.

(i) The probability of performing the routine correctly is 0.65, and incorrectly is 0.35. We use the binomial probability formula:

\(P(X = 5) = (0.65)^5 \times (0.35)^2 \times \binom{7}{5}\)

\(= (0.65)^5 \times (0.35)^2 \times 21\)

\(= 0.298\)

(iii) The expected number of correct performances is given by \(np\), where \(p = 0.65\). We need \(np \geq 8\).

\(0.65n \geq 8\)

\(n \geq \frac{8}{0.65} \approx 12.31\)

The smallest integer \(n\) is 13.

The probability that a tape is not damaged is 0.8 (since 1 in 5 are damaged, so 4 in 5 are not).

The probability that all n tapes in a sample are not damaged is given by (0.8)^n.

The probability that at least one tape is damaged is therefore 1 - (0.8)^n.

We need this probability to be at least 0.85:

1 - (0.8)^n \geq 0.85

(0.8)^n \leq 0.15

\(By trial and error or using logarithms, we find that n = 9 satisfies this inequality.\)

The probability that Janice does not buy an item online in any week is given by:

\(1 - 0.35 = 0.65\)

The probability that Janice does not buy any item online in n weeks is:

\((0.65)^n\)

The probability that Janice buys at least one item online in n weeks is greater than 0.99, so:

\(1 - (0.65)^n > 0.99\)

Rearranging gives:

\((0.65)^n < 0.01\)

Taking logarithms on both sides:

\(n \log(0.65) < \log(0.01)\)

Solving for n gives:

\(n > \frac{\log(0.01)}{\log(0.65)} \approx 10.69\)

Since n must be an integer, the smallest possible value of n is 11.

\(The probability that a customer does not shop online is 0.65 (since 1 - 0.35 = 0.65).\)

The probability that none of the n customers shop online is given by:

(0.65)^n

We want the probability that at least one customer shops online to be greater than 0.95:

1 - (0.65)^n > 0.95

Rearranging gives:

(0.65)^n < 0.05

Taking logarithms on both sides:

n \log(0.65) < \log(0.05)

Solving for n gives:

\(n > \frac{\log(0.05)}{\log(0.65)} \approx 6.95\)

Therefore, the least integer value of n is 7.