Answer: The radial component is \(\boxed{20\ \mathrm{m\ s^{-2}}}\), and the transverse component is \(\boxed{3\ \mathrm{m\ s^{-2}}}\).

The particle is moving on a circle of radius \(0.8\) m, and its speed at time \(t\) is

\[v=t^2-t+2.\]

At \(t=2\),

\[v=2^2-2+2=4.\]

The radial component of acceleration is

\[a_R=\frac{v^2}{r}.\]

So

\[a_R=\frac{4^2}{0.8}=\frac{16}{0.8}=20.\]

The transverse component is the rate of change of speed:

\[a_T=\frac{\mathrm dv}{\mathrm dt}=2t-1.\]

At \(t=2\),

\[a_T=2(2)-1=3.\]

Therefore the required magnitudes are \(\boxed{20\ \mathrm{m\ s^{-2}}}\) and \(\boxed{3\ \mathrm{m\ s^{-2}}}\).