Let A be the event that a motorist stopped at the first signal.
Let B be the event that a motorist stops at the second signal.
From the provided information, the probabilities are given as,
\(\begin{aligned}P\left( A \right) = 0.4\\P\left( B \right) = 0.5\\P\left( {A \cup B} \right) = 0.7\end{aligned}\)
a.
Using the formula for the probability of the union of the two events,
\(P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\)
The probability that he must stop at both signals is computed as,
\(\begin{aligned}P\left( {A \cap B} \right) &= P\left( A \right) + P\left( B \right) - P\left( {A \cup B} \right)\\ &= 0.4 + 0.5 - 0.7\\ &= 0.2\end{aligned}\)
Therefore, the probability that he must stop at both signals is 0.2.
b.
The probability that he stops at the first signal but not at the second one is computed as,
\(\begin{aligned}P\left( {A \cap B'} \right) &= P\left( A \right) - P\left( {A \cap B} \right)\\ &= 0.4 - 0.2\\ &= 0.2\end{aligned}\)
Therefore, the probability that he stops at the first signal but not at the second one is 0.2.
c.
The probability that he stops at the second signal but not at the first one is computed as,
\(\begin{aligned}P\left( {A' \cap B} \right) &= P\left( B \right) - P\left( {A \cap B} \right)\\ &= 0.5 - 0.2\\ &= 0.3\end{aligned}\)
Therefore, the probability that he stops at the second signal but not at the first one is 0.3.
The sum of the above computed probability and the probability computed in (b) part is the probability that he stops at exactly one signal.
The probability that he stops at exactly one signal is computed as,
\(\begin{aligned}P\left( {A \cap B'} \right) + P\left( {A' \cap B} \right) &= 0.2 + 0.3\\ &= 0.5\end{aligned}\)
Therefore, the probability that he stops at exactly one signal is 0.5.
