We use the equation for motion under constant acceleration: \(s = ut + \frac{1}{2}at^2\).

For the first 80 m in 5 s:

\(80 = 5u + \frac{1}{2}a \times 5^2\)

\(80 = 5u + 12.5a\)

For the first 160 m in 8 s:

\(160 = 8u + \frac{1}{2}a \times 8^2\)

\(160 = 8u + 32a\)

We now have two simultaneous equations:

1. \(5u + 12.5a = 80\)
2. \(8u + 32a = 160\)

Solving these equations, we first multiply the first equation by 8 and the second by 5 to eliminate \(u\):

\(40u + 100a = 640\)

\(40u + 160a = 800\)

Subtract the first from the second:

\(60a = 160\)

\(a = \frac{8}{3}\)

Substitute \(a = \frac{8}{3}\) back into the first equation:

\(5u + 12.5 \times \frac{8}{3} = 80\)

\(5u + \frac{100}{3} = 80\)

\(5u = 80 - \frac{100}{3}\)

\(5u = \frac{240}{3} - \frac{100}{3}\)

\(5u = \frac{140}{3}\)

\(u = \frac{28}{3}\)

Thus, \(a = \frac{8}{3} \text{ m s}^{-2}\) and \(u = \frac{28}{3} \text{ m s}^{-1}\).