As jameselmore points out in their comment, the Poisson distribution is more generally suitable for this kind of problem, where the arrival process is the juxtaposition of a large number of independent, individually rare occurrences (no bus arrives more than once). Additionally, one big appeal of the Poisson distribution is that it's "memoryless," meaning in this case that it doesn't matter how much you missed the previous bus by. Otherwise, you would have to take that into account.
And, when the Poisson distribution with rate $\lambda = 1/10$ counts the number of arrivals in a given interval of time, the interarrival time distribution is given by its partner in crime, the exponential distribution, with CDF
$$
F(t) \equiv P(T < t) = 1-e^{-\lambda t}
$$
In this case, you want $P(T \geq 15)$, which can be determined almost directly from the above as
$$
P(T \geq t) = 1-P(T < t) = 1-F(t) = e^{\lambda t}
$$
with $t = 15$. For part (2), you would then determine the probability that the Poisson distribution, with rate $\lambda = 1/10$ over interval $t = 5$ (note different $t$) yields $N \geq 2$ buses. The Poisson distribution is given by
$$
P(N = k) = \frac{(\lambda t)^k}{k!} e^{-\lambda t}
$$
and the desired probability is most easily determined as
$$
\begin{align}
P(N \geq 2) & = 1 - P(N = 0) - P(N = 1) \\
& = 1 - e^{-\lambda t} - \lambda t e^{-\lambda t}
\end{align}
$$
If you wanted to interpret the question as asking for the probability that exactly two buses (and no more) arrive within the five minutes after the first bus, you could determine that more directly as $P(N = 2)$.
So much for the Poisson/exponential distribution. Just for kicks, though, let's now suppose you were to use the uniform distribution. You would presumably use a distribution that was uniformly distributed over the interval $[0, 20]$. Part (2) doesn't make much sense unless it's possible for an interval to be less than $2 \frac{1}{2}$ minutes.
If that were the case, and interarrival times were independent (that seems odd, but let's assume that for the sake of brevity), and you just missed the last bus, then you could write
$$
P(\text{must wait at least $15$ minutes for next bus}) = 1-P(T_1 < 15)
= 1-\frac{15}{20} = \frac{1}{4}
$$
The probability that the second and third buses then show up within the next five minutes is probability that the sum of two interarrival times being less than five minutes. To do this, you would convolve the uniform distribution with itself, to get the PDF
$$
f(t) = \begin{cases}
\hfill \frac{t}{400} \hfill & 0 \leq t \leq 20 \\
\hfill \frac{40-t}{400} \hfill & 20 \leq t \leq 40
\end{cases}
$$
and we can then write
$$
\begin{align}
P(T_3-T_1 < 5) & = \int_{t=0}^5 f(t) \, dt \\
& = \frac{t^2}{800} \Bigr]_{t=0}^5 \\
& = \frac{25}{800} = \frac{1}{32}
\end{align}
$$
However, as we said, in general, the Poisson distribution makes more sense for this kind of example.
Best Answer
Hint: Let $B$ be the time (in minutes) elapsed since 10 o'clock. Then $B$ is uniformly distribute on the interval $[0,30]$ with density $\dfrac{1}{30}$.
The probability that the waiting time is longer than 10 minutes is $P(B ??)$.
If the bus hasn't arrived at 10:15 (we're given a condition, so think of the definition of conditional probability), the probability that one will have to wait for at least an additional 10 minutes is $P(B?? | B??)$.