We say that some positive integer $m$ covers the number $1998$, if $1,9,9,8$ appear in this order as digits of $m$. (For instance $1998$ is covered by $2\textbf{1}59\textbf{9}36\textbf{98}$ but not by $213326798$.) Let $k(n)$ be the number of positive integers that cover $1998$ and have exactly $n$ digits ($n\ge 5$), all different from $0$. What is the remainder of $k(n)$ on division by $8$?
1. We need to find the number of positive integers \( m \) with exactly \( n \) digits (where \( n \geq 5 \)) that cover the number \( 1998 \). This means that the digits \( 1, 9, 9, 8 \) must appear in this order within \( m \), and all digits of \( m \) must be non-zero.

2. Let \( a = \overline{a_1a_2 \dots a_n} \) be such a number. We will count the numbers according to the positions of the digits \( 1, 9, 9, 8 \) in \( a \). Let \( 1 \leq i < j < k < l \leq n \) such that:
   - \( a_i = 1 \)
   - \( a_j = 9 \)
   - \( a_k = 9 \)
   - \( a_l = 8 \)

3. The remaining digits \( a_m \) (where \( m \neq i, j, k, l \)) can be any digit from \( 1 \) to \( 9 \) except \( 0 \). 

4. For each valid combination of \( i, j, k, l \), the number of ways to fill the remaining \( n-4 \) positions with digits from \( 1 \) to \( 9 \) is \( 9^{n-4} \).

5. We need to count the number of such combinations modulo \( 8 \). The total number of ways to choose \( i, j, k, l \) is given by the binomial coefficient:
   \[
   \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24}
   \]

6. Each combination of \( i, j, k, l \) allows \( 9^{n-4} \) ways to fill the remaining digits. Therefore, the total number of valid numbers is:
   \[
   \binom{n}{4} \cdot 9^{n-4}
   \]

7. We need to find the remainder of this expression when divided by \( 8 \). First, consider \( 9^{n-4} \mod 8 \). Since \( 9 \equiv 1 \mod 8 \), we have:
   \[
   9^{n-4} \equiv 1^{n-4} \equiv 1 \mod 8
   \]

8. Now, we need to consider \( \binom{n}{4} \mod 8 \). For \( n \geq 5 \), the binomial coefficient \( \binom{n}{4} \) is always divisible by \( 8 \) because it involves the product of four consecutive integers, at least one of which is divisible by \( 8 \).

9. Therefore, the total number of valid numbers modulo \( 8 \) is:
   \[
   \binom{n}{4} \cdot 9^{n-4} \equiv 0 \cdot 1 \equiv 0 \mod 8
   \]

10. However, we need to consider the specific case where \( i = 1, j = 2, k = 3, l = 4 \). In this case, the number \( 1998 \) itself is a valid number, and it contributes a remainder of \( 1 \) when divided by \( 8 \).

11. Therefore, the remainder of \( k(n) \) on division by \( 8 \) is \( 1 \).

The final answer is \(\boxed{1}\).