Given R(2, 2) and S(5, 4), find the vector RS, its magnitude, and its bearing.

Given points $R(2, 2)$ and $S(5, 4)$.
The vector $\vec{RS}$ is found by subtracting the coordinates of $R$ from the coordinates of $S$.

$$\vec{RS} = S - R = (5-2, 4-2) = (3, 2)$$

The magnitude of a vector $(x, y)$ is given by the formula $\sqrt{x^2 + y^2}$.
Here, $x=3$ and $y=2$.

$$|\vec{RS}| = \sqrt{3^2 + 2^2}$$ $$|\vec{RS}| = \sqrt{9 + 4}$$ $$|\vec{RS}| = \sqrt{13}$$

The magnitude $k$ is $\sqrt{13}$.

The vector $\vec{RS} = (3, 2)$ is in the first quadrant.
Let $\alpha$ be the angle the vector makes with the positive x-axis (East).

$$\tan \alpha = \frac{\text{vertical component}}{\text{horizontal component}} = \frac{2}{3}$$ $$\alpha = \arctan\left(\frac{2}{3}\right)$$ $$\alpha \approx 33.69^\circ$$

Bearing is measured clockwise from the North direction (positive y-axis).
Since the vector is in the first quadrant, the bearing $\theta$ is $90^\circ - \alpha$.

$$\theta = 90^\circ - 33.69^\circ$$ $$\theta = 56.31^\circ$$

Rounding to one decimal place, $\theta \approx 56.3^\circ$.
As a 3-digit bearing, this is $056.3^\circ$.

From Step 2, $k = \sqrt{13}$.
From Step 3, $\theta \approx 056.3^\circ$.

i) The magnitude of $\vec{RS}$ is $\boxed{\sqrt{13}}$.

ii) Express $\vec{RS}$ in the form $(k, \theta)$: $\boxed{(\sqrt{13}, 056.3^\circ)}$.