(i) The acceleration \(a(t)\) is the derivative of the velocity \(v(t)\) with respect to time \(t\). Thus,

\(a(t) = \frac{dv}{dt} = \frac{d}{dt}(1.25t - 0.05t^2) = 1.25 - 0.1t\).

The initial acceleration is when \(t = 0\):

\(a(0) = 1.25 - 0.1 \times 0 = 1.25 \text{ m/s}^2\).

(ii) To find the displacement when the acceleration is \(0.05 \text{ m/s}^2\), set \(a(t) = 0.05\):

\(1.25 - 0.1t = 0.05\)

\(0.1t = 1.25 - 0.05\)

\(0.1t = 1.2\)

\(t = 12 \text{ s}\).

Now, integrate \(v(t)\) to find the displacement \(s(t)\):

\(s(t) = \int v(t) \, dt = \int (1.25t - 0.05t^2) \, dt\)

\(= \frac{1.25t^2}{2} - \frac{0.05t^3}{3} + C\).

Since the particle starts from rest, \(s(0) = 0\), so \(C = 0\).

Calculate \(s(12)\):

\(s(12) = \frac{1.25 \times 12^2}{2} - \frac{0.05 \times 12^3}{3}\)

\(= \frac{1.25 \times 144}{2} - \frac{0.05 \times 1728}{3}\)

\(= 90 - 28.8\)

\(= 61.2 \text{ m}\).