For part b I have found that the stationary state is $a_{n}=d/k$. For part $d)$ however I am not sure on how to take into account the statement 'immediately prior to the administration of the next dose'. Without that statement I would do $ d/k<1/2 $ in which I substituted $ d/k $ as the stationary state into $a_{n+1}$. However I am really not sure how to take that statement into account?
On
The idea is that the amount in the bloodstream is slowly decreasing between doses, then jumps up when a dose is administered. $a_n$ is defined as the amount in the bloodstream just after a dose, so is at the peak of the step. The amount just before dose $n$ is $a_n-d$. If $a$ is the stationary value, you need to make sure $a-d \ge \frac 12$
You should note the interpretation of the parameter $k$ and $d$. In this case, $k$ is the degradation rate of the drug in the blood stream. And $d$ is the amount of drug being added with each drug administration (not rate).
As @Did commented, the amount just before the dose (or prior to $a_{n+1}$) is equal to the amount from previous dose ($a_n$) subtracts total amount of drug being used up by the body ($a_n k$). Thus your condition is:
$$a_n - k a_n > \frac{1}{2}$$
Now, at equilibrium, $a_n = a_\infty = d/k$, so the condition becomes:
$$\frac{d}{k} - k \frac{d}{k} > \frac{1}{2}$$
Note that since $k$ is a rate, $k < 1$, so the above expression can be simplified to:
$$ d > \frac{1}{2} \frac{k}{1 - k}$$