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

Step 1: Calculate the volume of the hemispherical tank. The formula for the volume of a hemisphere is $V = \frac{2}{3} \pi r^3$. Given radius $r = 2 \text{ m}$ and $\pi = \frac{22}{7}$. $V = \frac{2}{3} \times \frac{22}{7} \times (2 m)^3$ $V = \frac{2}{3} \times \frac{22}{7} \times 8 m^3$ $V = \frac{352}{21} m^3$

Step 2: Convert the volume from cubic meters to litres. We know that $1 m^3 = 1000 \text{ litres}$. $V_{litres} = \frac{352}{21} \times 1000 \text{ litres}$ $V_{litres} = \frac{352000}{21} \text{ litres}$ $V_{litres} \approx 16761.90 \text{ litres}$

Step 3: Calculate the total water required for the farm. The farm area is $500 m^2$. Each square metre requires $50 \text{ litres}$ of water. Total water required = Farm area $\times$ Water per square metre. $\text{Water required} = 500 m^2 \times 50 litres/m^2$ $\text{Water required} = 25000 \text{ litres}$

Step 4: Compare the tank's capacity with the water required and state the conclusion. The tank capacity is approximately $16761.90 \text{ litres}$. The water required for the farm is $25000 \text{ litres}$. Since $16761.90 litres < 25000 \text{ litres}$, the full tank of water is not sufficient.

Mathematical statement: $V_{tank} < \text{Water required}$ $\frac{352000}{21} litres < 25000 \text{ litres}$ The full tank of water would $\boxed{not be sufficient}$ to irrigate the farm.