(a) Consider the expansion of \((2x^2 + \frac{a}{x})^6\). The general term is given by:

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

For \(x^6\), we have:

\(\binom{6}{2} (2x^2)^4 \left(\frac{a}{x}\right)^2 = 15 \times 16x^8 \times \frac{a^2}{x^2} = 15 \times 16a^2 x^6\)

For \(x^3\), we have:

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

Equating the coefficients:

\(15 \times 16a^2 = 20 \times 8a^3\)

\(240a^2 = 160a^3\)

\(a = \frac{3}{2}\)

(b) In the expansion of \((1-x^3)(2x^2 + \frac{a}{x})^6\), the term \(x^6\) comes from the product of \(-x^3\) and the constant term in \((2x^2 + \frac{a}{x})^6\), which is zero. Therefore, the coefficient of \(x^6\) is 0.

To find the coefficient of \(\frac{1}{x}\) in the expansion, we consider each part separately:

1. For \(\left( kx + \frac{1}{x} \right)^5\), the term that contributes to \(\frac{1}{x}\) is obtained by choosing 4 factors of \(kx\) and 1 factor of \(\frac{1}{x}\). The coefficient is:

\(\binom{5}{1} (kx)^4 \left( \frac{1}{x} \right)^1 = 5 \cdot k^4 \cdot x^3 \cdot \frac{1}{x} = 5k^4 x^2\)

2. For \(\left( 1 - \frac{2}{x} \right)^8\), the term that contributes to \(\frac{1}{x}\) is obtained by choosing 7 factors of 1 and 1 factor of \(-\frac{2}{x}\). The coefficient is:

\(\binom{8}{1} 1^7 \left( -\frac{2}{x} \right)^1 = 8 \cdot (-2) = -16\)

Combining these, the total coefficient of \(\frac{1}{x}\) is:

\(10k^2 - 16\)

We are given that this coefficient is 74:

\(10k^2 - 16 = 74\)

Solving for \(k\):

\(10k^2 = 90\)

\(k^2 = 9\)

\(k = 3\)

(a) To find the coefficient of \(\frac{1}{x}\), consider the term \(\binom{5}{2} (2x)^3 \left( \frac{a}{x^2} \right)^2\).

This simplifies to \(10 \times 8x^3 \times \frac{a^2}{x^4} = 720 \times \frac{1}{x}\).

Thus, \(80a^2 = 720\), giving \(a^2 = 9\).

Therefore, \(a = \pm 3\).

(b) To find the coefficient of \(\frac{1}{x^7}\), consider the term \(\binom{5}{4} (2x)^1 \left( \frac{a}{x^2} \right)^4\).

This simplifies to \(5 \times 2x \times \frac{a^4}{x^8}\).

Using \(a = 3\), the coefficient is \(5 \times 2 \times 81 = 810\).

First, expand \(\left(1 + \frac{x}{2}\right)^6\) using the binomial theorem:

\(\left(1 + \frac{x}{2}\right)^6 = 1 + \binom{6}{1}\frac{x}{2} + \binom{6}{2}\left(\frac{x}{2}\right)^2 + \cdots\)

The terms involving \(x\) and \(x^2\) are:

\(\binom{6}{1}\frac{x}{2} = 6\frac{x}{2} = 3x\)

\(\binom{6}{2}\left(\frac{x}{2}\right)^2 = 15\frac{x^2}{4}\)

Now, consider the expansion of \((4 + ax)(1 + 3x + 15\frac{x^2}{4})\).

The coefficient of \(x^2\) comes from:

\(4 \times 15\frac{x^2}{4} = 15x^2\)

\(ax \times 3x = 3ax^2\)

Equating the coefficient of \(x^2\) to 3:

\(3a + 15 = 3\)

Solving for \(a\):

\(3a = 3 - 15\)

\(3a = -12\)

\(a = -4\)

(i) To find the term independent of x, consider the general term in the expansion:

\(T_r = \binom{6}{r} (2x)^{6-r} \left( \frac{k}{x} \right)^r\)

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

\((6-r) - r = 0\)

\(6 - 2r = 0\)

\(r = 3\)

Substitute \(r = 3\) into the term:

\(T_3 = \binom{6}{3} (2x)^3 \left( \frac{k}{x} \right)^3\)

\(= 20 \cdot 8x^3 \cdot \frac{k^3}{x^3}\)

\(= 20 \cdot 8 \cdot k^3\)

\(= 160k^3\)

Given that this term is 540, we have:

\(160k^3 = 540\)

\(k^3 = \frac{540}{160}\)

\(k^3 = \frac{27}{8}\)

\(k = \frac{1}{2}\)

(ii) To find the coefficient of x², consider the term:

\(T_r = \binom{6}{r} (2x)^{6-r} \left( \frac{k}{x} \right)^r\)

We need \((6-r) - r = 2\), so:

\(6 - 2r = 2\)

\(4 = 2r\)

\(r = 2\)

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

\(T_2 = \binom{6}{2} (2x)^4 \left( \frac{k}{x} \right)^2\)

\(= 15 \cdot 16x^4 \cdot \frac{k^2}{x^2}\)

\(= 15 \cdot 16 \cdot k^2 \cdot x^2\)

Substitute \(k = \frac{1}{2}\):

\(= 15 \cdot 16 \cdot \left( \frac{1}{2} \right)^2\)

\(= 15 \cdot 16 \cdot \frac{1}{4}\)

\(= 240\)