(a) To find the speed, integrate the acceleration function. The acceleration is given by \(a = 4t^{\frac{1}{2}}\). Integrating with respect to \(t\), we have:

\(v = \int 4t^{\frac{1}{2}} \, dt = \frac{4}{1.5} t^{\frac{3}{2}} + c = \frac{8}{3} t^{\frac{3}{2}} + c\)

Since the particle starts from rest, \(v = 0\) when \(t = 0\), so \(c = 0\).

Substitute \(t = 9\) to find the speed:

\(v = \frac{8}{3} (9)^{\frac{3}{2}} = 72 \text{ m/s}\)

(b) To find when the speed and distance are equal, first find the distance by integrating the speed function:

\(s = \int \frac{8}{3} t^{\frac{3}{2}} \, dt = \frac{8}{3} \cdot \frac{2}{5} t^{\frac{5}{2}} + c = \frac{16}{15} t^{\frac{5}{2}} + c\)

Since \(s = 0\) when \(t = 0\), \(c = 0\).

Set \(v = s\) to find \(t\):

\(\frac{8}{3} t^{\frac{3}{2}} = \frac{16}{15} t^{\frac{5}{2}}\)

\(\frac{8}{3} = \frac{16}{15} t\)

\(t = \frac{5}{2} \text{ s}\)

Given the acceleration function: \(a(t) = 0.8t^{-0.75}\).

To find the velocity \(v(t)\), integrate the acceleration:

\(v(t) = \int 0.8t^{-0.75} \, dt = \frac{0.8}{0.25} t^{0.25} + C\)

\(C\) is the constant of integration. Given initial velocity \(v(0) = 1.8\), we find \(C = 1.8\).

Thus, \(v(t) = 3.2t^{0.25} + 1.8\).

To find the displacement \(s(t)\), integrate the velocity:

\(s(t) = \int (3.2t^{0.25} + 1.8) \, dt = \frac{3.2}{1.25} t^{1.25} + 1.8t + K\)

Since the particle starts from \(O\), \(s(0) = 0\), so \(K = 0\).

Thus, \(s(t) = 2.56t^{1.25} + 1.8t\).

Substitute \(t = 16\):

\(s(16) = 2.56(16)^{1.25} + 1.8(16)\)

\(s(16) = 111\)

Therefore, the displacement of \(P\) from \(O\) when \(t = 16\) is 111 m.

(i) To verify that \(t = 100\) when P is at A, substitute \(t = 100\) into the velocity equation:

\(v(100) = 0.16 \times 1000 - 0.016 \times 10000 = 0\).

Since \(v = 0\), P is at rest, confirming \(t = 100\).

(ii) To find the maximum speed, first find the acceleration \(a\) by differentiating \(v\):

\(a = \frac{dv}{dt} = 1.5 \times 0.16t^{\frac{1}{2}} - 0.032t\).

Set \(a = 0\) to find critical points:

\(1.5 \times 0.16t^{\frac{1}{2}} = 0.032t\).

Solving gives \(t^{\frac{1}{2}} = \frac{0.24}{0.032} \Rightarrow t = 56.25\).

Substitute \(t = 56.25\) into \(v(t)\) to find maximum speed:

\(v_{\text{max}} = 0.16 \times 421.875 - 0.016 \times 3164.0625 = 16.9 \text{ m s}^{-1}\).

(iii) To find the distance OA, integrate \(v\) with respect to \(t\):

\(s = \int v \, dt = \frac{2}{5} \times 0.16t^{\frac{5}{2}} - 0.016 \times \frac{t^3}{3}\).

Evaluate from \(t = 0\) to \(t = 100\):

\(s = 1070 \text{ m}\).

(iv) To find when P passes through O on returning, solve \(s(t) = 0\):

\(\frac{1}{3} t^{\frac{1}{2}} (0.192 - 0.016 \sqrt{t}) = 0\).

Solving gives \(t = 144\).

(i) To find the initial speed, integrate the acceleration to get the velocity function:

\(v(t) = \int a \, dt = \int \left( \frac{3}{160}t^2 - \frac{1}{800}t^3 \right) \, dt = \frac{1}{160}t^3 - \frac{1}{3200}t^4 + C_1\)

Given that the particle comes to rest at B (\(v(20) = 0\)), we have:

\(0 = \frac{1}{160}(20)^3 - \frac{1}{3200}(20)^4 + C_1\)

\(0 = 8000/160 - 160000/3200 + C_1\)

\(C_1 = 0\)

Thus, the initial speed \(v(0) = 0\), confirming the initial speed is zero.

(ii) To find the maximum speed, set the acceleration to zero and solve for \(t\):

\(a = \frac{3}{160}t^2 - \frac{1}{800}t^3 = 0\)

\(t^2(15 - t) = 0\)

\(t = 0\) or \(t = 15\)

The maximum speed occurs at \(t = 15\):

\(v(15) = \frac{1}{160}(15)^3 - \frac{1}{3200}(15)^4 = 5.27 \text{ m/s}\)

(iii) To find the distance AB, integrate the velocity function:

\(s(t) = \int v(t) \, dt = \int \left( \frac{1}{160}t^3 - \frac{1}{3200}t^4 \right) \, dt = \frac{1}{640}t^4 - \frac{1}{16000}t^5 + C_2\)

Using limits from 0 to 20:

\(s(20) - s(0) = \left( \frac{1}{640}(20)^4 - \frac{1}{16000}(20)^5 \right) - 0\)

\(= 250 - 200 = 50 \text{ m}\)

Thus, the distance AB is 50 m.

(i) To find the time taken, we need to integrate the velocity function to get the displacement function:

\(s = \int v \, dt = \int (4t - \frac{1}{16}t^3) \, dt = 2t^2 - \frac{t^4}{64} + C\)

Given that the distance PQ is 64 m, we set \(s = 64\):

\(2t^2 - \frac{t^4}{64} = 64\)

\(t^4 - 128t^2 + 64^2 = 0\)

\((t^2 - 64)^2 = 0\)

Solving gives \(t^2 = 64\), so \(t = 8\) s.

(ii) The acceleration is the derivative of the velocity:

\(a = \frac{dv}{dt} = \frac{d}{dt}(4t - \frac{1}{16}t^3) = 4 - \frac{3}{16}t^2\)

Set \(a > 0\):

\(4 - \frac{3}{16}t^2 > 0\)

\(\frac{3}{16}t^2 < 4\)

\(t^2 < \frac{64}{3}\)

\(t < \frac{8}{\sqrt{3}} \approx 4.62\)

Thus, the set of values for \(t\) is \(0 < t < \frac{8}{\sqrt{3}}\) or \(0 < t < 4.62\).