Find the velocity function.

The acceleration is given by $a = (4t - 13) \text{ ms}^{-2}$.
Velocity $v$ is the integral of acceleration with respect to time $t$.

We are given that the initial velocity is $18 \text{ ms}^{-1}$, which means $v = 18$ when $t = 0$.
Substitute these values to find $C$:

So, the velocity function is:

The particle is momentarily at rest when its velocity $v = 0$.

We can solve this quadratic equation by factorization. We need two numbers that multiply to $2 \times 18 = 36$ and add up to $-13$. These numbers are $-4$ and $-9$.

This gives two possible values for $t$:

The values of $t$ when the particle is momentarily at rest are $\boxed{\text{2 s and 4.5 s}}$.

Position $s$ is the integral of velocity with respect to time $t$.

For calculating distance covered, the constant $D$ will cancel out, so we can set $D=0$.

From part (a), we know the particle is momentarily at rest at $t = 2$ s and $t = 4.5$ s.
The value $t = 2$ s lies within the interval $[1, 3]$ s. This means the particle changes direction at $t=2$ s. Therefore, we must calculate the distance covered in two separate intervals: $[1, 2]$ s and $[2, 3]$ s, and sum their absolute displacements.

Displacement from $t=1$ to $t=2$:

Displacement $\Delta s_1 = s(2) - s(1) = \frac{46}{3} - \frac{73}{6} = \frac{92 - 73}{6} = \frac{19}{6} \text{ m}$.

Displacement from $t=2$ to $t=3$:

Displacement $\Delta s_2 = s(3) - s(2) = \frac{27}{2} - \frac{46}{3} = \frac{81 - 92}{6} = -\frac{11}{6} \text{ m}$.

Total distance covered is the sum of the absolute values of the displacements:

The distance covered by the particle between $t = 1$ second and $t = 3$ seconds is $\boxed{\text{5 m}}$.