It's perhaps a little more complex than it appears, requiring multiple steps.
First, let's define the train interarrival time ($x$) cumulative distribution function as $F(x)$, where $\int xf(x)dx = 7.5$. Now, it should be intuitively clear that given, say, two interarrival times $x_1$ and $x_2$, the probability that our hapless student arrives during interval $x_1$ is proportional to the length of $x_1$, and similarly for $x_2$. Since the probability of an interval being of length $x_1$ in the first place $ = f(x_1)$, we can see that:
$$p(\text{Observed interarrival time} = x) \propto xf(x)$$
Integrating $xf(x)$ to find the constant of proportionality (well, the part not already hidden in $f(x)$) gives us $7.5$, the mean interarrival time. So we have:
$$p(x) = {xf(x) \over 7.5}$$
(We will use that $7.5$ later on.) Now, if the student arrives randomly during an interval of length $x$, the arrival time is uniformly distributed over $(0,x)$, which of course implies the remaining time $t$ in the interval - i.e., the time until the next train arrives - is also distributed uniformly over $(0,x)$. Therefore, $p(t|x) = (1/x)\, 1(t<x)$, where $1(a)$ is the indicator function taking on the value $1$ if the condition $a$ is true, $0$ otherwise.
Combining the two expressions gives us:
$$p(t, x) = p(t|x)p(x) = {1 \over x}{xf(x) \over 7.5}1(t<x) = {1 \over 7.5}f(x)1(t<x)$$
Now we want to integrate out $x$ so we can get the marginal distribution of $t$. The indicator function makes it clear that the appropriate range of integration of $x$ for any given $t$ is from $t$ to $\infty$, as for $x < t$ the function being integrated will equal $0$.
$$p(t) = {1 \over 7.5}\int_t^{\infty}f(x)dx = {1 \over 7.5}(1-F(t))$$
A quick check: $\int_0^{\infty}(1-F(t))dt = \mathbb{E}[t] = 7.5$ (this is a moderately well-known relationship), so we have ${1 \over 7.5}\int_0^{\infty}(1-F(t))dt = 1$ and we have derived a proper probability distribution. (Of course, the more general solution is $p(t) = (1-F(t))/\mathbb{E}[t]$.)
I have ignored the issue of what happens if the student arrives exactly when the train leaves - does he get on the train, in which case the uniform distribution of the waiting time conditional on $x$ is over $[0, x)$, or does she have to wait for the next one, in which case it's $(0,x]$, with appropriate changes to the indicator function etc. Fortunately the difference amounts to a set of measure zero, i.e., the probability of that occurring equals zero, so our final result holds either way.
