(a) To find the term independent of x in \(\left( 3x + \frac{2}{x^2} \right)^6\), use the binomial expansion:

\(\binom{6}{k} (3x)^{6-k} \left( \frac{2}{x^2} \right)^k\).

We need the power of x to be zero:

\(6-k - 2k = 0 \Rightarrow 6 = 3k \Rightarrow k = 2\).

Substitute \(k = 2\):

\(\binom{6}{2} (3x)^4 \left( \frac{2}{x^2} \right)^2 = 15 \times 81x^4 \times \frac{4}{x^4} = 4860\).

(b) For \(\left( 3x + \frac{2}{x^2} \right)^6 (1 - x^3)\), find the term independent of x:

Use the term from part (a) and subtract the term where \(x^3\) is multiplied by the term in \(\left( 3x + \frac{2}{x^2} \right)^6\) that results in \(x^{-3}\).

Find the term in \(\left( 3x + \frac{2}{x^2} \right)^6\) with \(x^{-3}\):

\(6-k - 2k = -3 \Rightarrow 6 = 3k - 3 \Rightarrow k = 3\).

Substitute \(k = 3\):

\(\binom{6}{3} (3x)^3 \left( \frac{2}{x^2} \right)^3 = 20 \times 27x^3 \times \frac{8}{x^6} = 4320\).

Subtract this from the term in part (a):

\(4860 - 4320 = 540\).