Two people, P and Q, start at the same time from a base camp R. P walks on a bearing of 205° at 4 km/h and Q walks on a bearing of 060° at 3 km/h. Calculate the distance between P and Q after 5 hours.

P moves on a bearing of 205° from R. Q moves on a bearing of 060° from R. The angle $\angle PRQ$ between their paths from R is the difference between these bearings.

$$\angle PRQ = 205^\circ - 060^\circ = 145^\circ$$

We have a triangle RPQ with sides $RP = 20$ km, $RQ = 15$ km, and the included angle $\angle PRQ = 145^\circ$.

$$PQ^2 = RP^2 + RQ^2 - 2(RP)(RQ)\cos(\angle PRQ)$$ $$PQ^2 = 20^2 + 15^2 - 2(20)(15)\cos(145^\circ)$$ $$PQ^2 = 400 + 225 - 600(-0.819152)$$ $$PQ^2 = 625 + 491.4912$$ $$PQ^2 = 1116.4912$$ $$PQ = \sqrt{1116.4912} \approx 33.4139 \text{ km}$$

Rounding to the nearest whole number:

$$PQ \approx \boxed{\text{33 km}}$$

To find the bearing of Q from P, we first need to determine an angle within $\triangle RPQ$. Let's find $\angle RPQ$.

$$\frac{RQ}{\sin(\angle RPQ)} = \frac{PQ}{\sin(\angle PRQ)}$$ $$\frac{15}{\sin(\angle RPQ)} = \frac{33.4139}{\sin(145^\circ)}$$ $$\sin(\angle RPQ) = \frac{15 \sin(145^\circ)}{33.4139}$$ $$\sin(\angle RPQ) = \frac{15 \times 0.573576}{33.4139}$$ $$\sin(\angle RPQ) = \frac{8.60364}{33.4139} \approx 0.25747$$ $$\angle RPQ = \arcsin(0.25747) \approx 14.92^\circ$$

The bearing of P from R is 205°. The back bearing of R from P is $205^\circ - 180^\circ = 25^\circ$. This is the angle from the North line at P, measured clockwise to the line segment PR.
The bearing of Q from P is the angle from the North line at P, measured clockwise to the line segment PQ.

$$\text{Bearing of Q from P} = (\text{Back bearing of R from P}) + \angle RPQ$$ $$\text{Bearing of Q from P} = 25^\circ + 14.92^\circ$$ $$\text{Bearing of Q from P} = 39.92^\circ$$

Rounding to the nearest whole number and expressing as a three-figure bearing:

$$\text{Bearing of Q from P} \approx \boxed{040^\circ}$$