(i) To find when the particle comes to instantaneous rest, we need to find when the velocity v is zero. The velocity is the integral of acceleration:

\(v = \int (0.4t^3 - 4.8t^{1/2}) \, dt = 0.1t^4 - 3.2t^{3/2} + c\)

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

\(v = 0.1t^4 - 3.2t^{3/2}\)

\(Setting v = 0 gives:\)

\(0.1t^4 = 3.2t^{3/2}\)

Solving for t, we get:

\(t^{5/2} = 32\)

\(t = 4\)

\(Substitute t = 4 into the acceleration equation:\)

\(a = 0.4(4)^3 - 4.8(4)^{1/2} = 16\)

Thus, the acceleration is 16 m s^-2.

(ii) To find the displacement, integrate the velocity:

\(s = \int (0.1t^4 - 3.2t^{3/2}) \, dt = \left[0.02t^5 - 1.28t^{5/2}\right]_0^5\)

Calculate the displacement:

\(s = (0.02(5)^5 - 1.28(5)^{5/2}) - (0.02(0)^5 - 1.28(0)^{5/2})\)

\(s = -9.05\)

Thus, the displacement is -9.05 m.