The binomial expansion of \((2+ax)^n\) gives the first three terms as:

1st term: \(2^n\)

2nd term: \(n \cdot 2^{n-1} \cdot (ax)\)

3rd term: \(\frac{n(n-1)}{2} \cdot 2^{n-2} \cdot (ax)^2\)

Given: 1st term = 32, so \(2^n = 32\). Therefore, \(n = 5\).

2nd term: \(5 \cdot 2^4 \cdot ax = -40x\). Solving for \(a\), we get:

\(80a = -40\)

\(a = -\frac{1}{2}\)

3rd term: \(\frac{5 \cdot 4}{2} \cdot 2^3 \cdot (ax)^2 = bx^2\). Substituting \(a = -\frac{1}{2}\), we have:

\(10 \cdot 8 \cdot \left(-\frac{1}{2}\right)^2 = b\)

\(b = 20\)