Let’s determine the probability that a first request arrives in the next two seconds. First, note that since X is a continuous variable that f(x) is NOT the probability of exactly X minutes but you must integrate to compute all probabilities. Also, the next 2 seconds is actually the next \(\frac{2}{60} = \frac{1}{30}\) of a minute. Therefore, F(x) is what you need in general and you find
\begin{equation*}
P(X \le \frac{1}{30}) = F(\frac{1}{30}) = 1 - e^{-\frac{\frac{1}{30}}{0.01}} \approx 0.96433.
\end{equation*}