Relate acceleration to velocity.

Acceleration $a$ is the rate of change of velocity $V$ with respect to time $t$.

Given $a = (5 - 2t) \, \text{m/s}^2$.
So, we have:

The initial velocity is $2 \, \text{m/s}$, which means $V=2$ when $t=0$.

The expression for $V$ in terms of $t$ is $\boxed{V = -t^2 + 5t + 2 \, \text{m/s}}$.

The beginning of the third second corresponds to $t=2$ seconds.

The velocity at the beginning of the third second is $\boxed{8 \, \text{m/s}}$.

Maximum velocity occurs when the acceleration is zero ($a = \frac{dV}{dt} = 0$).
Given $a = 5 - 2t$.
Set $a = 0$:

The time taken to attain maximum velocity is $\boxed{2.5 \, \text{s}}$.

Distance $s$ is the integral of velocity $V$ with respect to time $t$.

Using $V = -t^2 + 5t + 2$:

Assume that at $t=0$, the displacement $s=0$.

So, the displacement equation is:

To simplify, use fractions: $t = \frac{5}{2}$.

Find a common denominator, which is 24:

The distance covered to attain the maximum velocity is $\boxed{\frac{185}{12} \, \text{m} \text{ or } 15.42 \, \text{m (to 2 d.p.)}}$.