I do not know how to do d) and e) but I think it has something to do with Central Limit Theorem and using Taylor Expansion. Any help would be much appreciated.
I have done some working and I think d) may correspond to a Normal distribution with $\text{mean} = 1$ and $\text{variance} = 0$, but am very unsure.
For (a) I got a Negative Binomial MGF $\mathrm{NegBin}(3 , 0.75)$ which came to $$M_{Y_3}(t)= \frac{(0.75e^t)^3}{(1-0.25e^t)^3}$$
for (b) I got $\mathrm{NegBin}(n , 0.75)$ $$M_{Y_n}(t)=\frac{(0.75e^t)^n}{(1-0.25e^t)^n}$$
for (c) I got an MGF of $$M_{\text{sample mean}}(t)=\frac{(0.75e^{t/n})^n}{(1-0.25e^{t/n})^n}$$