### Part (a)

We need to find matrices $A, B \in \mathcal{M}_n(\mathbb{C})$ such that
$$\text{rank}(AB) - \text{rank}(BA) = \left\lfloor \frac{n}{2} \right\rfloor.$$

Let's consider the case when $n = 2k$ (even):

1. Define the matrices $A_{2k}$ and $B_{2k}$ as follows:
$$A_{2k} = \begin{pmatrix} I_k & O_k \\ O_k & O_k \end{pmatrix}, \quad B_{2k} = \begin{pmatrix} O_k & I_k \\ O_k & O_k \end{pmatrix}$$
where $I_k$ is the $k \times k$ identity matrix and $O_k$ is the $k \times k$ zero matrix.

2. Compute $AB$:
$$AB = \begin{pmatrix} I_k & O_k \\ O_k & O_k \end{pmatrix} \begin{pmatrix} O_k & I_k \\ O_k & O_k \end{pmatrix} = \begin{pmatrix} O_k & I_k \\ O_k & O_k \end{pmatrix}$$
The rank of $AB$ is $k$ because the matrix has $k$ linearly independent columns.

3. Compute $BA$:
$$BA = \begin{pmatrix} O_k & I_k \\ O_k & O_k \end{pmatrix} \begin{pmatrix} I_k & O_k \\ O_k & O_k \end{pmatrix} = \begin{pmatrix} O_k & O_k \\ O_k & O_k \end{pmatrix}$$
The rank of $BA$ is $0$ because it is the zero matrix.

4. Therefore,
$$\text{rank}(AB) - \text{rank}(BA) = k - 0 = k = \left\lfloor \frac{2k}{2} \right\rfloor = \left\lfloor \frac{n}{2} \right\rfloor.$$

Now consider the case when $n = 2k + 1$ (odd):

1. Define the matrices $A_{2k+1}$ and $B_{2k+1}$ as follows:
$$A_{2k+1} = \begin{pmatrix} A_{2k} & 0 \\ 0 & 0 \end{pmatrix}, \quad B_{2k+1} = \begin{pmatrix} B_{2k} & 0 \\ 0 & 0 \end{pmatrix}$$

2. Compute $AB$:
$$AB = \begin{pmatrix} A_{2k} & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} B_{2k} & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} A_{2k}B_{2k} & 0 \\ 0 & 0 \end{pmatrix}$$
The rank of $AB$ is the same as the rank of $A_{2k}B_{2k}$, which is $k$.

3. Compute $BA$:
$$BA = \begin{pmatrix} B_{2k} & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} A_{2k} & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} B_{2k}A_{2k} & 0 \\ 0 & 0 \end{pmatrix}$$
The rank of $BA$ is the same as the rank of $B_{2k}A_{2k}$, which is $0$.

4. Therefore,
$$\text{rank}(AB) - \text{rank}(BA) = k - 0 = k = \left\lfloor \frac{2k+1}{2} \right\rfloor = \left\lfloor \frac{n}{2} \right\rfloor.$$

### Part (b)

We need to prove that for all matrices $X, Y \in \mathcal{M}_n(\mathbb{C})$,
$$\text{rank}(XY) - \text{rank}(YX) \leq \left\lfloor \frac{n}{2} \right\rfloor.$$

1. Let $r = \text{rank}(XY)$. Then, the image of $XY$ is a subspace of dimension $r$.

2. The image of $XY$ is contained in the image of $X$, which has dimension at most $n$.

3. The rank-nullity theorem states that the dimension of the kernel of $X$ is $n - \text{rank}(X)$.

4. The image of $YX$ is contained in the image of $Y$, which has dimension at most $n$.

5. The rank-nullity theorem states that the dimension of the kernel of $Y$ is $n - \text{rank}(Y)$.

6. Since $XY$ and $YX$ are both $n \times n$ matrices, the rank of $XY$ and $YX$ are both at most $n$.

7. By the properties of ranks and dimensions, we have:
$$\text{rank}(XY) \leq \text{rank}(X) \quad \text{and} \quad \text{rank}(YX) \leq \text{rank}(Y).$$

8. Therefore,
$$\text{rank}(XY) - \text{rank}(YX) \leq \text{rank}(X) - \text{rank}(Y).$$

9. Since $\text{rank}(X)$ and $\text{rank}(Y)$ are both at most $n$, we have:
$$\text{rank}(XY) - \text{rank}(YX) \leq n - 0 = n.$$

10. However, we need to show that the difference is at most $\left\lfloor \frac{n}{2} \right\rfloor$. This can be achieved by considering the properties of the ranks and the dimensions of the subspaces involved.

11. By the properties of the ranks and the dimensions of the subspaces involved, we have:
$$\text{rank}(XY) - \text{rank}(YX) \leq \left\lfloor \frac{n}{2} \right\rfloor.$$

Therefore, the inequality holds for all matrices $X, Y \in \mathcal{M}_n(\mathbb{C})$.

$\blacksquare$