(a) Use the formula for distance traveled in the nth second:

\(s = u + \frac{1}{2}a(2n-1)\).

For the 2nd second:

\(52 = u + \frac{1}{2}a(3)\).

For the 4th second:

\(64 = u + \frac{1}{2}a(7)\).

These simplify to:

\(52 = u + 1.5a\)

\(64 = u + 3.5a\).

Subtract the first equation from the second:

\(12 = 2a\)

\(a = 6 \text{ m/s}^2\).

Substitute \(a = 6\) into \(52 = u + 1.5a\):

\(52 = u + 9\)

\(u = 43 \text{ m/s}\).

(b) Use the formula for total distance traveled:

\(s = ut + \frac{1}{2}at^2\).

Substitute \(u = 43\), \(a = 6\), and \(t = 10\):

\(s = 43 \times 10 + \frac{1}{2} \times 6 \times 10^2\)

\(s = 430 + 300\)

\(s = 730 \text{ m}\).