I believe the first question should be $P(W_{20} - W_1 \le 10)$ since you do not care about the time before the first person arrives.
For the second, it may be useful to use results about Poisson superposition and thinning.
Instead of thinking about two independent Poisson processes (passengers and buses), it turns out you can think of a single Poisson process of rate $4+1=5$, and then for each arrival flip a biased coin to decide if it is a passenger arrival (probability $\frac{4}{5}$) or a bus arrival (probability $\frac{1}{5}$).
So, if you want the probability that exactly $n$ people arrive before the first bus arrives, then this is simply $(4/5)^n (1/5)$.
If you want the probability that at least $n$ people arrive before the first bus arrives, then it's $\sum_{k=n}^\infty (4/5)^k (1/5)$ which can be simplified using geometric series to be $(4/5)^n$.
Of course, this approach is contingent on knowing the aforementioned results about Poisson process.
Alternate approach: if you want to follow your approach, the probability that at least $n$ people arrive before the first bus arrives is indeed $P(W_n^{(p)} < W_1^{(b)})$. If you first condition on $W_1^{(b)} = t$, then
$$P(W_n^{(p)} < W_1^{(b)} \mid W_1^{(b)} = t) = P(W_n^{(p)} < t) = 1 - \sum_{k=0}^{n-1} e^{-4t} \frac{(4t)^k}{k!}.$$
Then indeed you can integrate as you have set it up.
$$P(W_n^{(p)} < W_1^{(b)}) = 1 - \sum_{k=0}^{n-1} \int_0^\infty e^{-4t} \frac{(4t)^k}{k!} \cdot e^{-t} \, dt
= 1- \sum_{k=0}^{n-1} (4/5)^k (1/5) \underbrace{\int_0^\infty e^{-5t} \frac{5^{k+1} t^k}{k!}\, dt}_{=1}.$$
This leads to the same answer as above.
Your answer is correct, so it remains to address the question of Why is the solution independent of $\lambda$?
This in turn is equivalent to asking why conditioning on $N=3$ arrivals in $[0,t]$,
the arrival times are uniformly distributed and independent of $\lambda$, as this is the only fact you rely on from the theory of Poisson processes.
You can either prove this for yourself (I suspect you have a proof given that you were aware of the result in your post), but doing so from the arrival process construct of Poisson processes doesn't give much intuition.
Instead I'll walk through the non-arrival time construct of the Poisson process and argue that from this perspective there's no intuition that the distribution of the points in an interval should depend on $\lambda$.
The spatial (non-time dependent) definition of the process is roughly as follows: for any interval $(a,b)$ the number of points that occur in the interval is Poisson distributed with mean $\lambda(b-a)$, and any two disjoint intervals are independent.
Now with this in mind, suppose we consider the intervals $(0,t/2)$ and $(t/2,t)$. From the definition above both are independent, and both will have $\text{Poi}(\lambda t/2)$ points.
If we now condition that there is a single point in $(0,t)$, then is it more likely this point falls into $(0,t/2)$ or $(t/2,t)$? From the preceding paragraph, a priori there's no reason to prefer one set over the other, eg. the point is equally likely to be in either - independently of $\lambda$.
Extrapolating out: we could equally consider the case of conditioning on multiple points $n$, and breaking an interval $(0,t)$ into an arbitrary number of sub-intervals, and we'd still conclude that any given point is no more likely to land in one of the boxes than any other. So informally we've concluded that the distribution of points is independent of $\lambda$, and uniformly distributed.
Best Answer
Your procedure to find $\mathbb E[Y\mid X(t)]$ is incorrect. You calculate the $\mathbb ES_i$ without taking into account that they must be calculated under a condition.
Let $U_1,\dots, U_n$ be iid rv's with uniform distribution on $[0,t]$ and let $U_{(1)},\dots, U_{(n)}$ be the corresponding order statistics.
Then it can be shown that under condition $X(t)=n$ the distribution of $(S_1,\dots,S_n)$ is the same as the distribution of $(U_{(1)},\dots, U_{(n)})$.
Consequently the distribution (and expectation) of $S_1+\cdots+S_n$ is the same as the distribution (and expectation) of $U_{(1)}+\cdots+U_{(n)}=U_1+\cdots+U_n$.
So if $Y$ denotes the total waiting time then $$\mathbb E[Y\mid X(t)=n]=\mathbb E[(t-S_1)+\cdots +(t-S_n)\mid X(t)=n]=\frac12nt$$
So: $$\mathbb E[Y\mid X(t)]=\frac12 X(t)t$$ and consequently:$$\mathbb EY=\mathbb E[\mathbb E[Y\mid X(t)]]=\mathbb E\frac12X(t)t=\frac12\lambda t^2$$