(a) To find the initial speed, use the equation of motion:

\(v^2 = u^2 + 2as\)

At the maximum height, \(v = 0\), \(a = -g\), and \(s = 45\) m. Thus,

\(0 = u^2 - 2g \times 45\)

Solving for \(u\), we get:

\(u^2 = 2g \times 45\)

\(u = \sqrt{2 \times 9.8 \times 45} = 30 \text{ m s}^{-1}\)

(b) To find the total time for which the speed is at least 10 m s^-1, use the equation:

\(v = u + at\)

Set \(v = 10 \text{ m s}^{-1}\), \(u = 30 \text{ m s}^{-1}\), and \(a = -g\):

\(10 = 30 - gt\)

\(gt = 20\)

\(t = \frac{20}{9.8} \approx 2 \text{ s}\)

The particle takes 2 s to reach 10 m s^-1 on the way up and another 2 s on the way down, so the total time is:

\(2 \times 2 = 4 \text{ s}\)