I have completed a) for question d)
I have attempted the following questions c and d and I am wondering what are the correct solutions and answers?
c) Is it reasonable to use Poisson model for this problem? Give the reasons.
d) Using Poisson model find the probability that more than two of the male births have Klinefelter syndrome (up to four decimal places) and taking binomial probability in (a) as the base (that is, 100%), by what percentage does the Poisson approximation differ from the binomial probability?
The Poisson approximation to the binomial is valid when $n$ is large and $p$ is small. In this case, $p = 0.0011$ and $n = 1000$, hence the approximation should be reasonable. Note that the Poisson approximation need not require $\lambda = np$ to be small.
Because you used the Poisson approximation to compute your answer to parts (a) and (b), rather than using the binomial distribution to compute the exact probability for those parts, the only way to answer part (d) now is to compute the exact probability and compare that to the value you got in (a). You want $$\Pr[X > 2] = 1 - \Pr[X \le 2] = 1 - \Pr[X = 0] - \Pr[X = 1].$$ You seem to have trouble interpreting which values of the random variable to select for the corresponding event; this was the same difficulty you had in computing (b) earlier.
Proceeding, we have the exact probability using the binomial distribution $$\Pr[X > 2] = 1 - \binom{n}{0} p^0 (1-p)^n - \binom{n}{1} p^1 (1-p)^{n-1},$$ and with $n = 1000$ and $p = 0.0011$, we get $$\begin{align*} \Pr[X > 2] &= 1 - (0.9989)^{1000} - 1000 (0.0011)(0.9989)^{999} \\ &= 1 - (0.9989)^{999} (0.9989 + 1.1) \\ &= 1 - (2.0989)(0.9989)^{999} \\ &\approx 0.300991. \end{align*}$$ The Poisson approximation gave you about $0.300971$, for a difference on the order of $0.00002$. Using the binomial probability as the base, the error of $0.00002$ is approximately $$\frac{0.00002}{0.300991} \times 100\% \approx 0.0067\%$$ of the true probability.