Determine the initial position components.

The aircraft is 60 km north-east from OR Tambo. This means it is at an angle of $45^\circ$ from the East (or North).

$$x_0 = 60 \cos(45^\circ) = 60 \times \frac{\sqrt{2}}{2} = 30\sqrt{2} \approx 42.426 \text{ km}$$ $$y_0 = 60 \sin(45^\circ) = 60 \times \frac{\sqrt{2}}{2} = 30\sqrt{2} \approx 42.426 \text{ km}$$

The aircraft flies north for 6 hours at a velocity of 96 km/h.

$$D_f = \text{velocity} \times \text{time} = 96 \text{ km/h} \times 6 \text{ h} = 576 \text{ km (North)}$$

The final position components are the sum of the initial position and the flight displacement.

$$x_{\text{final}} = x_0 = 42.426 \text{ km}$$ $$y_{\text{final}} = y_0 + D_f = 42.426 \text{ km} + 576 \text{ km} = 618.426 \text{ km}$$

The magnitude of the final position is:

$$R = \sqrt{x_{\text{final}}^2 + y_{\text{final}}^2} = \sqrt{(42.426)^2 + (618.426)^2}$$ $$R = \sqrt{1799.96 + 382450.6} = \sqrt{384250.56} \approx 619.88 \text{ km}$$

The direction of the final position is:

$$\theta = \arctan\left(\frac{y_{\text{final}}}{x_{\text{final}}}\right) = \arctan\left(\frac{618.426}{42.426}\right)$$ $$\theta = \arctan(14.575) \approx 86.06^\circ$$

The position is $86.06^\circ$ North of East.
The final answer is $\boxed{\text{619.88 km at } 86.06^\circ \text{ North of East}}$.

Let upward be the positive direction.
Velocity of hoist K (descending): $V_K = -6.48 \text{ km/h}$
Velocity of hoist L (ascending): $V_L = +5.04 \text{ km/h}$
The velocity of hoist K relative to hoist L is:

$$V_{KL} = V_K - V_L = -6.48 \text{ km/h} - (+5.04 \text{ km/h}) = -11.52 \text{ km/h}$$

The magnitude is $11.52 \text{ km/h}$ and the negative sign indicates a downward direction.
The final answer is $\boxed{\text{11.52 km/h downward}}$.

The velocity of hoist L relative to hoist K is:

$$V_{LK} = V_L - V_K = +5.04 \text{ km/h} - (-6.48 \text{ km/h}) = 5.04 \text{ km/h} + 6.48 \text{ km/h} = +11.52 \text{ km/h}$$

The magnitude is $11.52 \text{ km/h}$ and the positive sign indicates an upward direction.
The final answer is $\boxed{\text{11.52 km/h upward}}$.

Given: initial velocity $v_0 = 42 \text{ m/s}$, angle $\theta = 26^\circ$, $g = 9.8 \text{ m/s}^2$.
The formula for maximum height is:

$$H = \frac{(v_0 \sin \theta)^2}{2g}$$ $$H = \frac{(42 \sin 26^\circ)^2}{2 \times 9.8}$$ $$H = \frac{(42 \times 0.43837)^2}{19.6} = \frac{(18.4095)^2}{19.6}$$ $$H = \frac{338.909}{19.6} \approx 17.29 \text{ m}$$

The final answer is $\boxed{\text{17.29 m}}$.

The formula for horizontal range is:

$$R = \frac{v_0^2 \sin(2\theta)}{g}$$ $$R = \frac{(42)^2 \sin(2 \times 26^\circ)}{9.8}$$ $$R = \frac{1764 \sin(52^\circ)}{9.8}$$ $$R = \frac{1764 \times 0.78801}{9.8} = \frac{1390.07}{9.8} \approx 141.84 \text{ m}$$

The final answer is $\boxed{\text{141.84 m}}$.