First, find the coefficient of \(x^4\) in \((3 + x)^5\). Using the binomial theorem, the general term is \(\binom{5}{r} 3^{5-r} x^r\). For \(x^4\), set \(r = 4\):

\(\binom{5}{4} 3^{1} x^4 = 5 \times 3 = 15\).

Next, find the coefficient of \(x^2\) in \(\left(2x + \frac{a}{x}\right)^6\). The general term is \(\binom{6}{r} (2x)^{6-r} \left(\frac{a}{x}\right)^r\).

For \(x^2\), solve \(6-r-r = 2\), giving \(r = 2\).

The term is \(\binom{6}{2} (2x)^{4} \left(\frac{a}{x}\right)^2 = 15 \times 16x^4 \times \frac{a^2}{x^2} = 240a^2 x^2\).

Equating the coefficients: \(15 = 240a^2\).

Solving for \(a\):

\(a^2 = \frac{15}{240} = \frac{1}{16}\).

\(a = \frac{1}{4}\) (since \(a\) is positive).

(a) To find \(a\), consider the term in \(x^4\) from \(\left( 2x^2 + \frac{k^2}{x} \right)^5\). The term is given by:

\([10 \times x](2x^2)^3 \left( \frac{k^2}{x} \right)^2\)

\(80k^4 x^4 \Rightarrow a = 80k^4\).

To find \(b\), consider the term in \(x^2\) from \((2kx - 1)^4\). The term is given by:

\([6 \times 1](2kx)^2 \times 1 = 24k^2 x^2 \Rightarrow b = 24k^2\)

(b) Given \(a + b = 216\), substitute the expressions for \(a\) and \(b\):

\(80k^4 + 24k^2 = 216\)

\(10k^4 + 3k^2 - 27 = 0\)

Factorizing gives:

\((2k^2 - 3)(5k^2 + 9) = 0\)

\(k^2 = \frac{3}{2}\) or \(k^2 = -\frac{9}{5}\)

Since \(k^2\) must be non-negative, \(k^2 = \frac{3}{2}\).

Thus, \(k = \pm \frac{\sqrt{3}}{2}\).