(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):

\(4860 - 4320 = 540\).

To find the term independent of x in the expansion of \(\left( 2x + \frac{1}{4x^2} \right)^6\), we use the binomial theorem:

\(\binom{6}{k} (2x)^{6-k} \left( \frac{1}{4x^2} \right)^k\)

The term independent of x occurs when the powers of x cancel out, i.e., when:

\((6-k) - 2k = 0\)

Solving for k gives:

\(6 - 3k = 0\)

\(k = 2\)

Substitute \(k = 2\) into the binomial term:

\(\binom{6}{2} (2x)^4 \left( \frac{1}{4x^2} \right)^2\)

\(= 15 \times 16x^4 \times \frac{1}{16x^4}\)

\(= 15\)

Thus, the term independent of x is 15.

To find the term independent of x, we use the binomial expansion formula:

\(\left( a + b \right)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).

Here, \(a = 2x\) and \(b = -\frac{1}{4x^2}\), and \(n = 9\).

The general term is \(T_k = \binom{9}{k} (2x)^{9-k} \left(-\frac{1}{4x^2}\right)^k\).

Simplifying, \(T_k = \binom{9}{k} \cdot 2^{9-k} \cdot x^{9-k} \cdot (-1)^k \cdot \frac{1}{4^k} \cdot x^{-2k}\).

\(T_k = \binom{9}{k} \cdot 2^{9-k} \cdot (-1)^k \cdot \frac{1}{4^k} \cdot x^{9-k-2k}\).

\(T_k = \binom{9}{k} \cdot 2^{9-k} \cdot (-1)^k \cdot \frac{1}{4^k} \cdot x^{9-3k}\).

For the term to be independent of x, the power of x must be zero:

\(9 - 3k = 0\).

Solving for \(k\), we get \(k = 3\).

Substitute \(k = 3\) into the expression for \(T_k\):

\(T_3 = \binom{9}{3} \cdot 2^{9-3} \cdot (-1)^3 \cdot \frac{1}{4^3}\).

\(T_3 = \binom{9}{3} \cdot 2^6 \cdot (-1) \cdot \frac{1}{64}\).

\(T_3 = 84 \cdot 64 \cdot (-1) \cdot \frac{1}{64}\).

\(T_3 = -84\).

Thus, the term independent of x is \(-84\).

To find the term independent of x, we use the binomial expansion formula:

\(\left( a + b \right)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).

Here, \(a = 2x\) and \(b = \frac{1}{2x^3}\), and \(n = 8\).

We need the term where the power of x is zero. The general term is:

\(T_k = \binom{8}{k} (2x)^{8-k} \left( \frac{1}{2x^3} \right)^k\).

Simplifying, we get:

\(T_k = \binom{8}{k} \cdot 2^{8-k} \cdot x^{8-k} \cdot \frac{1}{2^k} \cdot x^{-3k}\).

\(T_k = \binom{8}{k} \cdot 2^{8-2k} \cdot x^{8-4k}\).

We want the power of x to be zero:

\(8 - 4k = 0\).

Solving for \(k\), we get \(k = 2\).

Substitute \(k = 2\) into the term:

\(T_2 = \binom{8}{2} \cdot 2^{8-4} \cdot x^{0}\).

\(T_2 = 28 \cdot 2^4 \cdot 1\).

\(T_2 = 28 \cdot 16\).

\(T_2 = 448\).

Thus, the term independent of x is 448.

To find the term independent of x, we use the binomial expansion formula:

\(\left( x - \frac{3}{2x} \right)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} \left( -\frac{3}{2x} \right)^k\)

The term independent of x occurs when the powers of x cancel out, i.e., when:

\(6-k-k = 0\)

Solving for k, we get:

\(6 - 2k = 0\)

\(k = 3\)

\(Substitute k = 3 into the binomial term:\)

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

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

\(= 20 \cdot x^3 \cdot \left( -\frac{3}{2x} \right)^3\)

\(= 20 \cdot x^3 \cdot \left( -\frac{27}{8x^3} \right)\)

\(= 20 \cdot \left( -\frac{27}{8} \right)\)

\(= -67.5\)