To solve this problem, we need to prove two statements about the composite integer $n$ in relation to the factorial $(n-k)!$.

### Part (a): If $n = 2p_k$, then $n$ does not divide $(n-k)!$

1. **Identify $n$ and $p_k$**:
Given $n = 2p_k$, where $p_k$ is the largest prime number strictly less than $k$.

2. Factorial and divisibility:
We need to show that $n = 2p_k$ does not divide $(n-k)!$.

3. Range of factorial:
Consider the factorial $(n-k)!$. Since $n = 2p_k$, we have:
$$n - k = 2p_k - k$$
The factorial $(2p_k - k)!$ includes all integers from 1 to $2p_k - k$.

4. **Prime $p_k$ in the factorial**:
Since $p_k$ is the largest prime less than $k$, and $p_k \ge \frac{3k}{4}$, it follows that:
$$2p_k - k \le 2 \left(\frac{3k}{4}\right) - k = \frac{3k}{2} - k = \frac{k}{2}$$
Therefore, $2p_k - k \le \frac{k}{2}$.

5. **Divisibility by $2p_k$**:
Since $2p_k - k \le \frac{k}{2}$, the factorial $(2p_k - k)!$ does not include the prime $p_k$ or the number $2p_k$. Hence, $2p_k$ cannot divide $(2p_k - k)!$.

Thus, $n = 2p_k$ does not divide $(n-k)!$.

### Part (b): If $n > 2p_k$, then $n$ divides $(n-k)!$

1. **Identify $n$ and $p_k$**:
Given $n > 2p_k$, where $p_k$ is the largest prime number strictly less than $k$.

2. Factorial and divisibility:
We need to show that $n$ divides $(n-k)!$.

3. Range of factorial:
Consider the factorial $(n-k)!$. Since $n > 2p_k$, we have:
$$n - k > 2p_k - k$$
The factorial $(n-k)!$ includes all integers from 1 to $n-k$.

4. **Inclusion of $p_k$ and $2p_k$**:
Since $n > 2p_k$, the range $n - k$ is greater than $2p_k - k$. Therefore, the factorial $(n-k)!$ includes both $p_k$ and $2p_k$.

5. **Divisibility by $n$**:
Since $n$ is a composite number greater than $2p_k$, and $(n-k)!$ includes all factors of $n$, it follows that $n$ divides $(n-k)!$.

Thus, $n > 2p_k$ divides $(n-k)!$.