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 \(a\):

\(3a = 3 - 15\)

\(3a = -12\)

\(a = -4\)