Let $N$ denote the number of cars that will be fully served, in other words, if $N=n$, then car $n+1$ is the first one ordering more petrol than the amount left at the station.
Changing slightly the notations and using the homogeneity of the problem,
$$
[N\geqslant n]=[\text{car}\ n\ \text{is fully served}]=[U_1+\cdots+U_n\leqslant2],
$$ where the random variables $U_n$ are i.i.d. and uniform on $(0,1)$. The mean time elapsed between two cars are served is $1/\lambda$ hence the mean time $\mathtt t$ before some petrol order cannot be satisfied is
$$
\mathtt t=\frac{E(N)+1}\lambda.
$$
Now, let $N_x$ denote the number of cars that will be fully served if the initial volume of petrol at the station is $x\cdot50$ liters. Then $N=N_2$ and, for every $x\geqslant0$, conditioning on the amount of petrol the first car is served and using the identity $P(U_1\leqslant x)=\min\{x,1\}$, one gets
$$
E(N_x)=\min\{x,1\}+\int_0^{\min\{x,1\}}E(N_{x-u})\,\mathrm du.
$$
Thus, if $x\geqslant1$ then
$$
E(N_x)=1+\int_0^1E(N_{x-u})\,\mathrm du=1+\int_{x-1}^xE(N_u)\,\mathrm du,
$$
while if $x\leqslant1$ then
$$
E(N_x)=x+\int_0^xE(N_{x-u})\,\mathrm du=x+\int_0^xE(N_u)\,\mathrm du.
$$
Let $n(x)=E(N_x)$. If $x\leqslant1$, then
$$
n'(x)=1+n(x),
$$
and $n(0)=0$ hence, for every $x$ in $(0,1)$,
$$
n(x)=\mathrm e^{x}-1.
$$
If $1\leqslant x\leqslant2$, then
$$
n'(x)=n(x)-n(x-1)=n(x)-(\mathrm e^{x-1}-1).
$$
Integrating this and using the initial condition $n(1)=\mathrm e-1$ yields, for every $x$ in $(1,2)$,
$$
n(x)=\mathrm e^x-1-\mathrm e^{x-1}(x-1),
$$
in particular,
$$
E(N)=n(2)=\mathrm e^2-1-\mathrm e,
$$
which yields
$$
\mathtt t=\frac{\mathrm e^2-\mathrm e}\lambda\approx\frac{4.67}\lambda.
$$
Your reasoning for (1) is incorrect. Because the process is Poisson, interarrival times are independent. So after the first car has arrived, subsequent arrivals occur as they would as if the process had just begun. But then, the third car to arrive is the second car we observe after the arrival of the first car. In other words, if the cars have arrival times $T_1$, $T_2$, $T_3$, then $T_3$ corresponds to the second arrival time after $T_1$. Thus, we must compute the probability that $T_3 - T_1 \le 10$, or equivalently, that at least two additional cars are observed in the 10 minutes following the observation of the first car.
Best Answer
Let the arrival process of red cars is according to the Poisson process, say, $\{R(t),t\ge 0\}$ with an intensity parameter $\lambda_1>0$ and the arrival process of blue cars is according to the Poisson process, say, $\{B(t),t\ge 0\}$ with an intensity parameter $\lambda_2>0$. Further, both the processes are independent of each other. We are interested in the number of arrivals of blue cars between two successive arrivals of red cars.
We know that, the inter-arrival times of a Poisson process with parameter $\lambda$ are exponential with mean $1/\lambda$. This means that, the pdf of inter arrival times between successive red cars is $$f(t)=\lambda_1 e^{-\lambda_1 t}.$$ Now, the probability distribution of $X$, the number of blue cars during an arbitrary interval (as determined by successive arrivals of red cars) is \begin{eqnarray*} P\{X=k\}&=&\int_{0}^{\infty}\dfrac{e^{-\lambda_2 t}(\lambda_{2}t)^{k}}{k!} f(t)dt\\ &=&\int_{0}^{\infty}\dfrac{e^{-\lambda_2 t}(\lambda_{2}t)^{k}}{k!} \left(\lambda_1 e^{-\lambda_1 t}\right) dt\\ &=&\dfrac{\lambda_1 \lambda_{2}^{k}}{k!}\int_{0}^{\infty}e^{-(\lambda_1 +\lambda_2)t}\cdot t^{(k+1)-1}dt\\ &=&\dfrac{\lambda_1 \lambda_{2}^{k}}{k!} \cdot \dfrac{\Gamma (k+1)}{(\lambda_1 +\lambda_2)^{k+1}}\\ P\{X=k\}&=&\left(\dfrac{\lambda_1}{\lambda_1 +\lambda_2}\right)\left(\dfrac{\lambda_2}{\lambda_1 +\lambda_2}\right)^{k},\quad k=0,1,2,\cdots \end{eqnarray*} which is a geometric distribution with parameter $\dfrac{\lambda_1}{\lambda_1 +\lambda_2}$.