Solve for P, the rate of ascent of a balloon, and the area of a sector.

$$P = 180^\circ - 140^\circ$$ $$P = 40^\circ$$

The size of angle P is \boxed{\text{40^\circ}}.

:
The balloon rises at a steady rate.

$$\text{Rate} = \frac{\text{Height}}{\text{Time}}$$ $$\text{Rate} = \frac{350 \text{ m}}{7 \text{ minutes}}$$ $$\text{Rate} = 50 \text{ m/minute}$$

$$\text{Height after 15 minutes} = \text{Rate} \times \text{Time}$$ $$\text{Height after 15 minutes} = 50 \text{ m/minute} \times 15 \text{ minutes}$$ $$\text{Height after 15 minutes} = 750 \text{ m}$$

The balloon will be $\boxed{\text{750 m}}$ high after 15 minutes.

:
The radius of the circular garden is $21 \text{ cm}$ and the central angle of the sector is $120^\circ$.

$$\text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2$$

where $\theta$ is the central angle and $r$ is the radius.

$$\text{Area of Sector} = \frac{120^\circ}{360^\circ} \times \pi (21 \text{ cm})^2$$ $$\text{Area of Sector} = \frac{1}{3} \times \pi (441 \text{ cm}^2)$$ $$\text{Area of Sector} = 147\pi \text{ cm}^2$$

Using $\pi \approx 3.14159$:

$$\text{Area of Sector} \approx 147 \times 3.14159 \text{ cm}^2$$ $$\text{Area of Sector} \approx 461.81 \text{ cm}^2$$

The area of the sector is approximately $\boxed{\text{461.81 cm}^2}$.

:
The equation of the straight line is $y = 4x - 8$.

$$y = 4(0) - 8$$ $$y = 0 - 8$$ $$y = -8$$

The coordinates of the point where the line cuts the y-axis are $\boxed{\text{(0, -8)}}$.