Generalization the time elapsed in Poisson Process

Suppose that people immigrate into a territory at a Poisson rate λ = 1 per day.

What is the probability that the elapsed time between the fourth and the sixth arrival exceeds two days, and What is the expected arrival time of the tenth immigrant given that exactly two immigrants arrived before and including the fifth day?

For the first part, I write

$$ \begin {align} &\mathbb P\{T_5 + T_6 >2\} \\ &= \mathbb P\{T_5>2\}\mathbb P\{T_6>0\}+\mathbb P\{T_5>0\}\mathbb P\{T_6>2\} + \mathbb P\{T_5 > 1 \} \mathbb P\{T_6 >1\} \\ &= e^{-2} + e^{-2} + 2e^{-1} \end {align} $$

And the second part, my approach is:

$$ \begin {align} & \mathbb E[S_{10} | S_{5}=2] \\&= \mathbb E[S_{10} | S_5 = 2]\mathbb P\{S_5=2\} \end {align} $$

I know that $\mathbb P\{S_5 = 2\}$ is nothing but a poisson with mean 5. But I have no idea on obtaining $E[S_5 = 8]$.

I have taken reference on Poisson process. Time between two events. But this post is related to 2 event only but not 3, I wondering that can I generalize it as well.

Poisson increments are independent, so the probability that the elapsed time between the $4$-th and $6$-th arrival is $>2$ equals the probability that the elapsed time of the $2$-nd arrival is $>2$ so we define $$\tau_2=\inf\{t\geq 0:N_t=2\}$$ and we want to find $P(\tau_2 >2)$. We have $$P(\tau_2>2)=P(N_2<2)=P(N_2=0)+P(N_2=1)=e^{-2}+2e^{-2}$$ For the second question we need to find $$E[\tau_{10}|N_5=2]$$ but this means finding the expected valued of the $8$-th arrival time, plus $5$. $$P(\tau_8>t)=P(N_t<8)=e^{-t}\sum_{k=0}^7\frac{t^k}{k!}$$ therefore $$E[\tau_8]=\int_{(0,\infty)}P(\tau_8>t)dt=\sum_{k=0}^7\frac{\Gamma(k+1)}{k!}=8$$ and $$E[\tau_{10}|N_5=2]=13$$