Given radius r = 2 m and pi = (22)/(7).

The formula for the volume of a hemisphere is $V = \frac{2}{3} \pi r^3$.
Given radius $r = 2 \text{ m}$ and $\pi = \frac{22}{7}$.

$$V = \frac{2}{3} \times \frac{22}{7} \times (2 \text{ m})^3$$ $$V = \frac{2}{3} \times \frac{22}{7} \times 8 \text{ m}^3$$ $$V = \frac{352}{21} \text{ m}^3$$

We know that $1 \text{ m}^3 = 1000 \text{ litres}$.

$$V_{litres} = \frac{352}{21} \times 1000 \text{ litres}$$ $$V_{litres} = \frac{352000}{21} \text{ litres}$$ $$V_{litres} \approx 16761.90 \text{ litres}$$

The farm area is $500 \text{ m}^2$.
Each square metre requires $50 \text{ litres}$ of water.
Total water required = Farm area $\times$ Water per square metre.

$$\text{Water required} = 500 \text{ m}^2 \times 50 \text{ litres/m}^2$$ $$\text{Water required} = 25000 \text{ litres}$$

The tank capacity is approximately $16761.90 \text{ litres}$.
The water required for the farm is $25000 \text{ litres}$.
Since $16761.90 \text{ litres} < 25000 \text{ litres}$, the full tank of water is not sufficient.

Mathematical statement:

$$V_{tank} < \text{Water required}$$ $$\frac{352000}{21} \text{ litres} < 25000 \text{ litres}$$

The full tank of water would $\boxed{\text{not be sufficient}}$ to irrigate the farm.