(i) The tree diagram is drawn with branches for stopping (S) and not stopping (NS) at each set of lights. The probabilities are labeled as follows: first light (S: 0.4, NS: 0.6), second light (S: 0.8, NS: 0.2), third light (S: 0.3, NS: 0.7).

(ii) The probability that Karinne stops at the first two sets but not the third is calculated as:

\(P(S, S, NS) = 0.4 \times 0.8 \times 0.7 = 0.224\) or \(\frac{28}{125}\).

(iii) The probability that Karinne stops at exactly two sets of lights is the sum of the probabilities of the following scenarios:

\(P(S, S, NS) + P(S, NS, S) + P(NS, S, S)\).

Calculating each:

\(P(S, S, NS) = 0.4 \times 0.8 \times 0.7 = 0.224\)

\(P(S, NS, S) = 0.4 \times 0.2 \times 0.3 = 0.024\)

\(P(NS, S, S) = 0.6 \times 0.8 \times 0.3 = 0.144\)

Summing these gives:

\(0.224 + 0.024 + 0.144 = 0.392\) or \(\frac{49}{125}\).

(iv) The probability that Karinne stops at the first set of lights, given that she stops at exactly two sets, is calculated using conditional probability:

\(P(S \text{ at first light} | \text{stops at exactly 2 lights}) = \frac{P(S, NS, S) + P(S, S, NS)}{0.392}\).

\(= \frac{0.024 + 0.224}{0.392} = \frac{0.248}{0.392} = 0.633\) or \(\frac{31}{49}\).