To find the kinetic energy of the body, we can use the formula:

$$KE = \frac{1}{2}mv^2$$

Given the mass $$m = (5 \pm 0.5) \,\text{kg}$$ and the velocity $$v = (20 \pm 0.4) \,\text{m/s}$$, we can find the kinetic energy and its uncertainty by applying the rules for the propagation of errors in multiplication.

For the product of two quantities, the relative error is the sum of the relative errors of the individual quantities:

$$\frac{\Delta (mv^2)}{mv^2} = \frac{\Delta m}{m} + \frac{\Delta v}{v} + \frac{\Delta v}{v}$$

Now, substitute the given values:

$$\frac{\Delta (mv^2)}{mv^2} = \frac{0.5}{5} + \frac{0.4}{20} + \frac{0.4}{20}$$

$$\frac{\Delta (mv^2)}{mv^2} = 0.1 + 0.02 + 0.02$$

$$\frac{\Delta (mv^2)}{mv^2} = 0.14$$

Now, calculate the kinetic energy:

$$KE = \frac{1}{2}(5)(20)^2 = 1000 \,\text{J}$$

To find the uncertainty in the kinetic energy, multiply the relative error by the calculated kinetic energy:

$$\Delta KE = 0.14 \times 1000 = 140 \,\text{J}$$

So, the kinetic energy of the body is $$(1000 \pm 140) \,\text{J}$$.