To solve the given problem, we need to understand the sequence $\{u_n\}_{n \in \mathbb{N}_0}$ defined by the following rules:
- $u_0 = 0$
- $u_{2n} = u_n$
- $u_{2n+1} = 1 - u_n$

We will use these rules to find $u_{1998}$ and $u_m$ where $m = (2^p - 1)^2$.

### Part (a): Finding $u_{1998}$

1. Binary Representation of 1998:
- Convert 1998 to its binary form:
$$1998 = 2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^3 + 2^2 + 2^1 = (11111001110)_2$$
- The binary representation of 1998 is $11111001110_2$.

2. Count the Number of 1's:
- Count the number of 1's in the binary representation:
$$\text{Number of 1's} = 7 \text{ (which is odd)}$$

3. **Determine $u_{1998}$**:
- According to the given rule, $u_n = 1$ if there is an odd number of 1's in the binary representation of $n$, and $u_n = 0$ if the number is even.
- Since 1998 has an odd number of 1's, we have:
$$u_{1998} = 1$$

### Part (b): Finding $u_m$ where $m = (2^p - 1)^2$

1. **Expression for $m$**:
- Given $m = (2^p - 1)^2$, we need to find its binary representation.
- Note that $2^p - 1$ in binary is a sequence of $p$ 1's:
$$2^p - 1 = 111\ldots1_2 \text{ (p times)}$$
- Squaring this number results in a binary number with a specific pattern.

2. **Binary Representation of $m$**:
- The binary representation of $m$ will have a block of $p-1$ 1's, followed by $p$ 0's, and ending with a single 1:
$$m = 2^{2p-1} + 2^{2p-2} + \ldots + 2^{p+2} + 2^{p+1} + 1 = (11\ldots1100\ldots01)_2$$
- This representation has $p$ 1's in total.

3. **Determine $u_m$**:
- Count the number of 1's in the binary representation of $m$:
$$\text{Number of 1's} = p$$
- According to the rule, $u_n = 1$ if there is an odd number of 1's in the binary representation of $n$, and $u_n = 0$ if the number is even.
- Therefore, $u_m = 0$ if $p$ is even, and $u_m = 1$ if $p$ is odd.

The final answer is: