(a) \[v=\frac{ds}{dt}=12t^2-78t+120=6(2t-5)(t-4).\] Setting \(v=0\) gives \(\boxed{t=2.5\text{ s},\ 4\text{ s}}\).

(b) The particle reverses direction at \(t=2.5\). Its displacements are \(s(1)=85\), \(s(2.5)=118.75\) and \(s(4)=112\). Add the distances in the two directions: \[D=(118.75-85)+(118.75-112)=\boxed{40.5\text{ m}}.\]

(c) \[a=\frac{dv}{dt}=24t-78,\qquad a(3)=72-78=-6\text{ m s}^{-2}.\] The requested magnitude is \(\boxed{6\text{ m s}^{-2}}\).