Given: Angle X = 35^, Angle Y = 65^, Side opposite X (side YZ) = 8 cm.

$$
\angle X + \angle Y + \angle Z = 180^\circ
$$
$$
35^\circ + 65^\circ + \angle Z = 180^\circ
$$
$$
100^\circ + \angle Z = 180^\circ
$$

$$
\angle Z = 180^\circ - 100^\circ
$$
$$
\angle Z = 80^\circ
$$
The measure of angle $Z$ is:
$$
\boxed{80^\circ}
$$

**b)** Find side $YZ$ using the sine rule.  
The sine rule states: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.  
We are given side $x = 8$ cm and $\angle X = 35^\circ$. We need to find side $YZ$, which is side $x$. The question asks to find side $YZ$ using the sine rule, but $YZ$ is already given as 8 cm. It is likely that the question meant to ask for side $XZ$ (side $y$) or side $XY$ (side $z$). Assuming it meant to find side $XZ$ (side $y$):

$$
\frac{x}{\sin X} = \frac{y}{\sin Y}
$$
$$
\frac{8 \text{ cm}}{\sin 35^\circ} = \frac{y}{\sin 65^\circ}
$$

$$
y = \frac{8 \text{ cm} \times \sin 65^\circ}{\sin 35^\circ}
$$
$$
y = \frac{8 \times 0.9063}{0.5736}
$$
$$
y = \frac{7.2504}{0.5736}
$$
$$
y \approx 12.64 \text{ cm}
$$
Assuming the question meant to find side $XZ$:
$$
\boxed{XZ \approx 12.64 \text{ cm}}
$$
If the question literally meant to find side $YZ$, then $YZ = 8$ cm, as given.

**c)** Find the area of the triangle.

$$
\text{Area} = 4 \times 12.64 \times 0.9848
$$
$$
\text{Area} = 4 \times 12.451
$$
$$
\text{Area} \approx 49.80 \text{ cm}^2
$$
The area of the triangle is:
$$
\boxed{49.80 \text{ cm}^2}
$$

**a)** Represent the journey on a scale diagram (1 cm = 20 km).

First leg: $120 \text{ km} \div 20 \text{ km/cm} = 6 \text{ cm}$.  
Second leg: $80 \text{ km} \div 20 \text{ km/cm} = 4 \text{ cm}$.

Let the starting point be A, the intermediate point B, and the final point C.  
The bearing from A to B is $030^\circ$.  
The bearing from B to C is $120^\circ$.  
The interior angle at B, $\angle ABC$, can be found. The angle from the North line at B to BA (back bearing) is $30^\circ + 180^\circ = 210^\circ$. The angle from the North line at B to BC is $120^\circ$.  
The angle between the North line at B and the line segment BA (pointing back to A) is $30^\circ$.  
The angle between the North line at B and the line segment BC is $120^\circ$.  
The angle $\angle ABC$ is the angle between BA and BC.  
The angle between the North line at B and the line BA (pointing towards A) is $30^\circ$.  
The angle between the South line at B and the line BA is $30^\circ$.  
The angle between the North line at B and the line BC is $120^\circ$.  
The angle between the South line at B and the line BC is $180^\circ - 120^\circ = 60^\circ$.  
The angle $\angle ABC = 30^\circ + 60^\circ = 90^\circ$.  
Alternatively, the interior angle at B is $180^\circ - (30^\circ + (180^\circ - 120^\circ)) = 180^\circ - (30^\circ + 60^\circ) = 180^\circ - 90^\circ = 90^\circ$.  
So, $\triangle ABC$ is a right-angled triangle.  
Side AB = 120 km, Side BC = 80 km.

$$
AC^2 = AB^2 + BC^2
$$
$$
AC^2 = (120 \text{ km})^2 + (80 \text{ km})^2
$$
$$
AC^2 = 14400 \text{ km}^2 + 6400 \text{ km}^2
$$
$$
AC^2 = 20800 \text{ km}^2
$$
$$
AC = \sqrt{20800} \text{ km}
$$
$$
AC \approx 144.22 \text{ km}
$$
The final displacement from the starting point is:
$$
\boxed{144.22 \text{ km}}
$$

**c)** Find the bearing of the final position from the starting point.

$$
\tan(\angle BAC) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB}
$$
$$
\tan(\angle BAC) = \frac{80 \text{ km}}{120 \text{ km}} = \frac{2}{3}
$$
$$
\angle BAC = \arctan\left(\frac{2}{3}\right)
$$
$$
\angle BAC \approx 33.69^\circ
$$

The initial bearing was $030^\circ$. The angle $\angle BAC$ is measured from the line AB towards AC.  
The bearing of C from A is $30^\circ + \angle BAC$.
$$
\text{Bearing} = 30^\circ + 33.69^\circ = 63.69^\circ
$$
Expressed as a three-figure bearing:
$$
\boxed{063.69^\circ}
$$

For $y = x$:
| $x$ | $y$ |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 2 |
| 3 | 3 |

For $x + y = 4$ (which can be rewritten as $y = 4 - x$):
| $x$ | $y$ |
|---|---|
| 0 | 4 |
| 1 | 3 |
| 2 | 2 |
| 3 | 1 |

**(Graph cannot be drawn here, but the steps describe how to construct it.)**

The two lines intersect at the point where their $x$ and $y$ values are the same. From the tables, we can see that both lines pass through the point $(2,2)$.  
Graphically, the intersection point will be $(2,2)$.  
The solution is:
$$
\boxed{x = 2, y = 2}
$$