Find the third angle of the triangle.

The sum of angles in a triangle is $180^\circ$.

$$\angle ACB = 180^\circ - \angle ABC - \angle CAB$$ $$\angle ACB = 180^\circ - 80^\circ - 45^\circ$$ $$\angle ACB = 55^\circ$$

The Sine Rule states $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. We want to find BC (side $a$).

$$\frac{BC}{\sin(\angle CAB)} = \frac{AC}{\sin(\angle ABC)}$$ $$\frac{BC}{\sin(45^\circ)} = \frac{50}{\sin(80^\circ)}$$

$$BC = \frac{50 \times \sin(45^\circ)}{\sin(80^\circ)}$$ $$BC = \frac{50 \times 0.7071}{0.9848}$$ $$BC = \frac{35.355}{0.9848}$$ $$BC \approx 35.900$$

The length of BC is approximately $\boxed{\text{35.90 units}}$.

2. In the triangle PQR, RPQ = 60, QRP = 50 and RP = 6.5. Find PQ.

$$\angle PQR = 180^\circ - \angle RPQ - \angle QRP$$ $$\angle PQR = 180^\circ - 60^\circ - 50^\circ$$ $$\angle PQR = 70^\circ$$

We want to find PQ (side $r$).

$$\frac{PQ}{\sin(\angle QRP)} = \frac{RP}{\sin(\angle PQR)}$$ $$\frac{PQ}{\sin(50^\circ)} = \frac{6.5}{\sin(70^\circ)}$$

$$PQ = \frac{6.5 \times \sin(50^\circ)}{\sin(70^\circ)}$$ $$PQ = \frac{6.5 \times 0.7660}{0.9397}$$ $$PQ = \frac{4.979}{0.9397}$$ $$PQ \approx 5.298$$

The length of PQ is approximately $\boxed{\text{5.30 units}}$.