(i) To find the time T when the particle reaches its maximum velocity, set the acceleration \(a = 0\):

\(1.2T^{1/2} - 0.6T = 0\)

Solving for \(T\):

\(1.2T^{1/2} = 0.6T\)

\(T^{1/2} = 0.5T\)

\(T^{1/2} = 2\)

\(T = 4\)

(ii) To find the velocity when the acceleration is maximum, first differentiate \(a\) with respect to \(t\):

\(\frac{da}{dt} = 0.6t^{-1/2} - 0.6\)

Set \(\frac{da}{dt} = 0\) to find \(t\):

\(0.6t^{-1/2} = 0.6\)

\(t^{-1/2} = 1\)

\(t = 1\)

Integrate \(a\) to find the velocity \(v\):

\(v = \int (1.2t^{1/2} - 0.6t) \, dt\)

\(v = 0.8t^{3/2} - 0.3t^2 + C\)

Using the initial condition \(v(0) = 1\):

\(1 = 0.8(0)^{3/2} - 0.3(0)^2 + C\)

\(C = 1\)

Thus, \(v = 0.8t^{3/2} - 0.3t^2 + 1\)

Substitute \(t = 1\) to find the velocity:

\(v = 0.8(1)^{3/2} - 0.3(1)^2 + 1\)

\(v = 0.8 - 0.3 + 1\)

\(v = 1.5\) m s^-1