Solution:

A counterexample for a) is $k=3$, $A=\{1,2,9\}$, $x=2$ and $d=8$.
A counterexample for c) is $k=3$, $A=\{1,8,9\}$, $x=8$ and $d=1$.

We will prove that b) is true.

Suppose the contrary and let $x, d$ have the above properties. We can assume $0 < d < 3k$, $0 < x \leq 3k$ (since for $d=3k$ the remainders for $x$ and $x+d$ are equal). Hence $0 < x+d < 6k$ and there are two cases:

- If $x+d > 3k$, then since the remainder for $x+d$ is medium we have $4k < x+d \leq 5k$. This means that the remainder of $x+d$ when it is divided by $3k$ is
$$x+d-3k$$
Since $x$ is medium we have $x \leq 2k$ so $d = (x+d) - x > 2k$. Therefore $6k = 4k + 2k < (x+d) + d < 8k$. This means that the remainder of $x+2d$ when it is divided by $3k$ is
$$x+2d-6k.$$
Thus the remainders $(x+2d-6k)$, $(x+d-3k)$ and $x$ are in $[1,3k]$, they belong to $A$ and
$$2(x+d-3k) = (x+2d-6k) + x$$
a contradiction.

- If $x+d \leq 3k$ then as $x+d$ is medium we have $k < x+d \leq 2k$. From the limitations on $x$, we have $x > k$ so $d = (x+d) - x < k$. Hence $0 \leq x+2d = (x+d) + d < 3k$. Thus the remainders $x, x+d$ and $x+2d$ are in $A$ and
$$2(x+d) = (x+2d) + x$$
a contradiction.