Given Information
Given that small aircraft arrive at a certain airport according to a Poisson process with rate $\alpha=8$ per hour, so that the number of arrivals during a time period of $t$ hours is a Poisson rv with parameter $\mu=8 t$.

Step-1: Part (a)
Probability that exactly 6 small aircraft arrive during a 1-hour period
$$\begin{aligned}P(X=6) &=p(6 ; 8) \\&=e^{-8} \dfrac{8^{6}}{6 !} \\&=0.122\end{aligned}$$Probability that at least 6 small aircraft arrive during a 1-hour period
$$\begin{aligned}P(X \geq 6) &=1-P(X<6) \\&=1-\sum_{x=0}^{5} p(x ; 8) \\&=1-\sum_{x=0}^{5} e^{-8} \dfrac{8^{x}}{x !} \\&=0.809\end{aligned}$$Probability that at least 10 small aircraft arrive during a 1-hour period
$$\begin{aligned}P(X \geq 10) &=1-P(X<10) \\&=1-\sum_{x=0}^{9} p(x ; 8) \\&=1-\sum_{x=0}^{9} e^{-8} \dfrac{8^{x}}{x !} \\&=0.283\end{aligned}$$

Step-2: Part (b)
t is required to find the expected value and standard deviation of the number of small aircraft that arrive during a 90-min period
The mean is
$$\begin{aligned}\mu_{Y} &=\mu \\&=12\end{aligned}$$The Standard deviation is $$\begin{aligned}\sigma_{Y} &=\sqrt{\mu} \\&=\sqrt{12} \\&=3.464\end{aligned}$$

Step-3: Part (c)
This is a poisson distribution with mean $$\begin{aligned}\mu &=2.5 \cdot 8 \\&=20\end{aligned}$$The probability that at least 20 small aircraft arrive during a 2.5-hour period
$$\begin{aligned}P(Z \geq 20) &=1-P(Z<20) \\&=1-\sum_{z=0}^{19} p(z ; 20) \\&=1-\sum_{z=0}^{19} e^{-20} \dfrac{20^{z}}{z !} \\&=0.530\end{aligned}$$The probability that at most 10 arrive during this period
$$\begin{aligned}P(Z \leq 10) &=\sum_{z=0}^{10} p(z ; 20) \\&=\sum_{z=0}^{10} e^{-20} \dfrac{20^{z}}{z !} \\&=0.011\end{aligned}$$==============================================================================================================