The question is this:
You arrive at a bus stop at 10 A.M., knowing that the bus will arrive at some time uniformly distributed between 10 and 10:30.
(a) What is the probability that you will have to wait longer than 10 minutes
The answer is the integral from 10 to 30 of 1/30dx. I don't understand why it is the integral of 1/30.
(b) If, at 10:15, the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes?
The answer is p(X>25|X>15) which I understand however I don't understand why the answer is then p(X>25)/p(x>15).
As you mentioned in the setup of the problem, the bus will arrive at some time uniformly between 10 and 10:30, so your wait time is therefore uniformly distributed between 0 and 30 minutes. Consequently, the wait time $X$ has the pdf $f(x) = \frac{1}{30}$ for $x \in [0, 30]$. Calculating the probability that we wait longer than 10 minutes, we then have $$P \left( X \geq 10 \right) = \int_{10}^{30} \frac{1}{30} dx = \frac{2}{3}$$
For part (b), you just need to note that the event $\{ X > 25 \} \subseteq \{ X > 15\}$ because if you wait longer than 25 minutes, then you have already waited longer than 15 minutes. Thus, we have the following: $\{ X > 25 , X > 15\} = \{X > 25\}$. Then we can use the definition of conditional probability to obtain the desired expression: $$P \left( X > 25 | X > 15 \right) = \frac{P \left(X > 25 , X > 15 \right)}{P \left( X > 15 \right)} = \frac{P \left( X > 25 \right)}{P \left( X > 15 \right)}$$