Q:

Crossing a lane of traffic

A pedestrian wishes to cross a single lane of fast-moving tra c. Suppose the number of vehicles that have passed by time t is a Poisson process of rate , and suppose it takes time a to walk across the lane. Assuming that the pedestrian can foresee correctly the times at which vehicles will pass by, 
Question: 1
how long on average does it take to cross over safely? [Consider the time at which the 1st car passes.]
Question 2: 
How long on average does it take to cross two similar lanes (a) when one must walk straight across (assuming that the pedestrian will not cross if, at any time whilst crossing, a car would pass in either direction), (b) when an island in the middle of the road makes it safe to stop half-way?

Attempt: 
Question 1:
It can be calculated by conditioning the first arrival. $E[X]=\int E[X|Y=y]f_Y(y)dy$, where $Y$ is the time of the first traffic. The answer is $(e^{\lambda a}-1)\lambda^{-1}$.
Question 2: 
a) I think this is two indepedent poisson process each with parameter $\lambda$. Therefore, the sum of the two poisson process = $2\lambda$. So $E[X]=(e^{2\lambda a}-1)(2\lambda)^{-1}$.
b)I think the answer is $2*(e^{\lambda a}-1)\lambda^{-1}$

A:

Question $1$: If a vehicle arrives before time $a$, you have to wait for it and then start over. Otherwise it takes you time $a$ to cross. The probability for the first vehicle to arrive at $t$ is $\lambda\mathrm e^{-\lambda t}$, so the expected time to cross is
$$
T=\int_0^a\lambda\mathrm e^{-\lambda t}(t+T)\mathrm dt+\int_a^\infty\lambda\mathrm e^{-\lambda t}a\mathrm dt=\frac1\lambda(1-\mathrm e^{-\lambda a})+T(1-\mathrm e^{-\lambda a})\;.
$$
Solving for $T$ yields $T=\frac1\lambda(\mathrm e^{\lambda a}-1)$, as you wrote.
Your answer for 2a) is almost right, but $a$ is also doubled, so the expected crossing time in this case is $\frac1{2\lambda}(\mathrm e^{4\lambda a}-1)$.
Your answer to 2b) is right and follows by linearity of expectation.

