The mean per page is $2.5$, so the mean per $30$ pages is $(30)(2.5)$.
This really has nothing much to do with the Poisson. Let $X_1$ be the number of typos on the first page, $X_2$ the number of typos on the second, and so on up to $30$. The total number $T$ of typos is given by
$$T=X_1+X_2+\cdots+X_{30}.$$
The mean of a sum is always the sum of the means ("linearity of expectation"). It follows that $E(T)=30E(X_1)=75$.
Where the Poisson comes in is not for the computation of the mean. But it is important for the distribution of $T$. Since the sum of independent Poissons is Poisson, $T$ has Poisson distribution.
According to the statement of the question, the sample that is collected is only representative of those individuals who experienced at least three alarms: therefore, the salient question is, what is the distribution of the sample mean under the assumption of the null hypothesis? That is, what is the distribution of $\bar X \mid H_0$ given that $X_i \ge 3$ for each $i = 1, 2, \ldots, 100$?
To this end, we observe that if $Y_i = (X_i \mid X_i \ge 3)$, $$\Pr[Y = x] = \frac{\Pr[(X_i = x) \cap (X_i \ge 3)]}{\Pr[X_i \ge 3]} = \frac{e^{-\lambda} \lambda^x/x!}{1-e^{-\lambda}(1+\lambda+\lambda^2/2)}, \quad x \ge 3.$$ Under the null hypothesis that $\lambda = 1$, this simply becomes $$\Pr[Y_i = x] = \frac{2}{(2e-5)x!}.$$ The expected value is $$\mu = \operatorname{E}[Y] = \sum_{x=3}^\infty x \Pr[Y_i = x] = 1 + \frac{1}{2e-5} \approx 3.29062.$$ The variance is $$\sigma^2 = \operatorname{Var}[Y] = \frac{2(7-8e+2e^2)}{(2e-5)^2} \approx 0.334309.$$ Therefore, the expected value of the sample mean under the null hypothesis is $$\mu_0 = \mu = 3.29062,$$ and the variance of the sample mean under the null hypothesis is $$\sigma_0^2 = \frac{\sigma^2}{100} = 0.00334309.$$ Using the Central Limit Theorem, $$\bar X \mid H_0 \, \dot\sim \operatorname{Normal}(\mu_0, \sigma_0^2),$$ hence we can perform a $z$-test: if $$Z = \frac{\bar X - \mu_0}{\sigma_0}$$ exceeds the $99^{\rm th}$ percentile of the standard normal distribution, then the sample contains sufficient evidence to suggest that the true value of $\lambda$ is strictly greater than $1$ with at most a Type I error probability of $\alpha = 0.01$.
Note that you cannot use the calculation you performed for part (a). The reason should now be clear: such a calculation would only be applicable when the sample is drawn from all people, not just those who observed at least 3 alarms. What we can see from the correct calculation is that if the mean number of alarms among those who had at least 3 is more than $3.42512$, then you would reject the one-sided test at the 1% significance level. The sample just isn't extreme enough.
As for part (b), I don't understand the intent here. As you can see, we are easily able to perform a meaningful hypothesis test with $n = 100$, which is large enough to justify the use of a normal approximation to the sampling distribution. The test statistic that you used, which seems to be implied by the question, is not the correct one to apply for the given hypothesis.
Best Answer
Imagine we have a population of fixed size $100000$, say a small city. The "two year $\lambda$" is $2.4$. Then the "one year $\lambda$" is $1.2$.
You can think of it as follows. If the mean number of occurrences of the disease in a $2$ year period is $2.4$, then the mean number of occurrences in a $1$ year period is $1.2$.
Remarks: $1.$ Suppose we now change the city size to say $300000$. Then the appropriate $\lambda$ for a $1$ year period becomes $3.6$.
$2.$ For finishing your problem about at least five, it is easier to first find the probability that there are $4$ or fewer occurrences. This is $e^{-\lambda}\left(1+\lambda+\frac{\lambda^2}{2!}+\frac{\lambda^3}{3!}+\frac{\lambda^4}{4!}\right)$.