In Fig. 6.52, the bearings of R from P and Q are 330° and 240°, respectively. If |PQ| = 50 km, calculate |QR| and the bearing Q from R.

The bearing of Q from P is $075^\circ$.
The bearing of R from P is $310^\circ$.
The angle between the lines PQ and PR, measured clockwise from PQ to PR, is $310^\circ - 075^\circ = 235^\circ$. This is the reflex angle.
The interior angle $\angle QPR$ is $360^\circ - 235^\circ = 125^\circ$.

In $\triangle PQR$, we have two sides (PQ and PR) and the included angle ($\angle QPR$). We can use the Cosine Rule:

Substitute the known values:

We know that $\cos(125^\circ) = -\cos(180^\circ - 125^\circ) = -\cos(55^\circ)$.

Rounding to two decimal places:
The distance between Q and R is $\boxed{\text{125.55 km}}$.

Now we need to find an angle within $\triangle PQR$ to help determine the bearing. Let's find $\angle PQR$.
Using the Sine Rule:

We know $\sin(125^\circ) = \sin(180^\circ - 125^\circ) = \sin(55^\circ)$.

First, find the back bearing of P from Q.
The bearing of Q from P is $075^\circ$.
The back bearing of P from Q is $075^\circ + 180^\circ = 255^\circ$. This means the line segment QP points in the direction $255^\circ$ from the North line at Q.

Now, we need to determine if $\angle PQR$ is added to or subtracted from the bearing of P from Q.
From P, Q is at $075^\circ$ and R is at $310^\circ$. This means R is to the "right" (clockwise) of the line PQ when viewed from P (since $310^\circ > 075^\circ$).
Therefore, when standing at Q and looking towards P (bearing $255^\circ$), R will be to your "right" (clockwise) relative to the line QP.
So, the bearing of R from Q will be the bearing of P from Q plus $\angle PQR$.

Rounding to the nearest degree (or two decimal places as calculated):
The bearing of R from Q is \boxed{\text{290.96^\circ}}.