Let $X_t$ be the number of requests that are received during $t$-hour period. And let $P_k(t)$ be the probability that $k$ requests are received during a particular $t$-hour period.
\[
P_{k}(t)=e^{-\lambda}\cdot\frac{(\lambda)^k}{k!}=e^{-\alpha t}\cdot\frac{(\alpha t)^k}{k!},
\]
here $\alpha=4$.
(a) For $k=10$ and $t=2$. We have $\lambda=\alpha t=4\times 2=8$.
\[
P(X_2=10)=P_{10}(2)=e^{-8}\cdot\frac{8^{10}}{10!}\approx 0.099262
\]
Hence, the probability that exactly ten requests are received during a particular $2$-hour period is about 0.099262.
(b)
If the operators of the towing service take a $30$-min break for lunch, the event that they do not miss any calls for assistance is equivalent to the event that there are zero requests during a $0.5$-hour period. That is, in this case,
$\lambda=\alpha\cdot t=4\times 0.5=2$.
Hence, the probability that they do not miss any calls for assistance is
\[
P_{0}(0.5)=e^{-\lambda}\cdot\frac{\lambda^0}{0!}=e^{-2}\cdot\frac{1}{1}=0.135335
\]
(c)
By proposition, the expected value of the number of requests during a $0.5$-hour period is $\mu=E(X_{0.5})=\lambda=2$.
That is, we expect 2 calls during for assistance their break.
