(a) To find when the particle is at instantaneous rest, set the velocity \(v = 0\). First, integrate the acceleration to find the velocity:

\(v = \int (36 - 6t) \, dt = 36t - 3t^2 + c\)

Given \(v = 27\) when \(t = 2\), substitute to find \(c\):

\(27 = 36(2) - 3(2)^2 + c\)

\(27 = 72 - 12 + c\)

\(c = -33\)

Thus, the velocity equation is:

\(v = 36t - 3t^2 - 33\)

Set \(v = 0\) to find \(t\):

\(0 = 36t - 3t^2 - 33\)

\(3t^2 - 36t + 33 = 0\)

Solving the quadratic equation gives \(t = 1\) and \(t = 11\).

(b) To find the total distance traveled, integrate the velocity to find the displacement:

\(s = \int (36t - 3t^2 - 33) \, dt = 18t^2 - t^3 - 33t + c\)

Evaluate from \(t = 0\) to \(t = 1\), \(t = 1\) to \(t = 11\), and \(t = 11\) to \(t = 12\):

\(s(0 \to 1) = 18(1)^2 - (1)^3 - 33(1) = -16\)

\(s(1 \to 11) = [18(11)^2 - (11)^3 - 33(11)] - [-16] = 500\)

\(s(11 \to 12) = [18(12)^2 - (12)^3 - 33(12)] - [18(11)^2 - (11)^3 - 33(11)] = -16\)

Total distance = \(|16| + 500 + |16| = 532 \text{ m}\).