To find the coefficient of \(x^3\) in the expansion of \((1 - px)^5\), we use the binomial theorem:

\((1 - px)^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (-px)^k\).

The term containing \(x^3\) is when \(k = 3\):

\(\binom{5}{3} (1)^{5-3} (-px)^3 = \binom{5}{3} (-1)^3 p^3 x^3\).

\(\binom{5}{3} = 10\), so the coefficient of \(x^3\) is:

\((-1) \cdot 10 \cdot p^3 = -10p^3\).

We are given that this coefficient is \(-2160\):

\(-10p^3 = -2160\).

Dividing both sides by \(-10\):

\(p^3 = 216\).

Taking the cube root of both sides:

\(p = 6\).

The coefficient of \(x^3\) in the expansion of \((3 + 2ax)^5\) is given by:

\(\binom{5}{3} (3)^{2} (2ax)^{3} = 10 \times 9 \times 8a^3 = 720a^3\)

The coefficient of \(x^2\) in the expansion of \((2 + ax)^6\) is given by:

\(\binom{6}{2} (2)^{4} (ax)^{2} = 15 \times 16 \times a^2 = 240a^2\)

According to the problem, the coefficient of \(x^3\) is six times the coefficient of \(x^2\):

\(720a^3 = 6 \times 240a^2\)

Simplifying gives:

\(720a^3 = 1440a^2\)

Dividing both sides by \(240a^2\) gives:

\(3a = 6\)

Thus, \(a = 2\).

First, find the coefficient of \(x^2\) in \(\left( 2 + \frac{x}{2} \right)^6\).

Using the binomial theorem, the term containing \(x^2\) is given by:

\(\binom{6}{2} \cdot 2^4 \cdot \left( \frac{x}{2} \right)^2 = 15 \cdot 16 \cdot \frac{x^2}{4} = 60x^2\)

So, the coefficient of \(x^2\) in \(\left( 2 + \frac{x}{2} \right)^6\) is 60.

Next, find the coefficient of \(x^2\) in \((a + x)^5\).

Using the binomial theorem, the term containing \(x^2\) is given by:

\(\binom{5}{2} \cdot a^3 \cdot x^2 = 10a^3x^2\)

So, the coefficient of \(x^2\) in \((a + x)^5\) is \(10a^3\).

According to the problem, the sum of these coefficients is 330:

\(60 + 10a^3 = 330\)

Solve for \(a^3\):

\(10a^3 = 270\)

\(a^3 = 27\)

\(a = 3\)

To find the coefficients of x and x² in the expansion of (2 + ax)⁷, we use the binomial theorem.

The general term in the expansion is given by:

\(\binom{7}{r} (2)^{7-r} (ax)^r\)

\(For the coefficient of x, set r = 1:\)

\(\binom{7}{1} (2)^6 (ax) = 7 \times 2^6 \times a \times x\)

The coefficient is \(7 \times 2^6 \times a\).

\(For the coefficient of x², set r = 2:\)

\(\binom{7}{2} (2)^5 (ax)^2 = 21 \times 2^5 \times a^2 \times x^2\)

The coefficient is \(21 \times 2^5 \times a^2\).

Since the coefficients are equal, we have:

\(7 \times 2^6 \times a = 21 \times 2^5 \times a^2\)

Simplifying gives:

\(7 \times 2 = 21 \times a\)

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

To find the coefficients of x² and x³ in the expansion of (3 - 2x)⁶, we use the binomial theorem:

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

For the term involving x², we have:

\(a = \binom{6}{2} \cdot 3^{4} \cdot (-2x)^{2}\)

\(a = 15 \cdot 81 \cdot 4\)

\(a = 4860\)

For the term involving x³, we have:

\(b = \binom{6}{3} \cdot 3^{3} \cdot (-2x)^{3}\)

\(b = 20 \cdot 27 \cdot (-8)\)

\(b = -4320\)

Thus, \(\frac{a}{b} = \frac{4860}{-4320} = -\frac{9}{8}\).

Therefore, the value of \(\frac{a}{b}\) is \(\frac{9}{8}\).