(i) Let the probability of a screw being faulty be \(p\). Given the mean number of faulty screws per packet is 1.2, we have:

\(1.2 = 15p\)

Solving for \(p\), we get:

\(p = \frac{1.2}{15} = 0.08\)

The variance \(\text{Var} = npq\), where \(q = 1 - p\).

\(\text{Var} = 15 \times 0.08 \times 0.92 = 1.104\)

(ii) The probability that a packet contains at most 2 faulty screws is:

\(P(0, 1, 2) = (0.92)^{15} + 15C_1(0.08)(0.92)^{14} + 15C_2(0.08)^2(0.92)^{13}\)

Calculating each term:

\((0.92)^{15} = 0.275\)

\(15 \times 0.08 \times (0.92)^{14} = 0.359\)

\(15C_2 \times (0.08)^2 \times (0.92)^{13} = 0.253\)

Adding these probabilities gives:

\(P(0, 1, 2) = 0.275 + 0.359 + 0.253 = 0.887\)

(iii) The probability that there is at least 1 faulty screw in a packet is:

\(P(\text{at least 1 faulty screw}) = 1 - P(0) = 1 - (0.92)^{15}\)

\(= 1 - 0.275 = 0.7137\)

The probability that there are exactly 7 packets with at least 1 faulty screw is:

\(P(\text{exactly 7 packets}) = 8C_7(0.7137)^7(0.2863)\)

\(= 0.216\)