(a) Use the binomial expansion formula: \((1 + ax)^n = 1 + nax + \frac{n(n-1)}{2!}a^2x^2 + \ldots\)

For \((1 + 2x)^5\):

First term: \(1\)

Second term: \(5 \times 2x = 10x\)

Third term: \(\frac{5 \times 4}{2} \times (2x)^2 = 40x^2\)

So, the first three terms are \(1 + 10x + 40x^2\).

(b) For \((1 - 3x)^4\):

First term: \(1\)

Second term: \(-4 \times 3x = -12x\)

Third term: \(\frac{4 \times 3}{2} \times (3x)^2 = 54x^2\)

So, the first three terms are \(1 - 12x + 54x^2\).

(c) To find the coefficient of \(x^2\) in \((1 + 2x)^5(1 - 3x)^4\), consider the products that result in \(x^2\):

1. \(1 \times 54x^2 = 54x^2\)

2. \(10x \times -12x = -120x^2\)

3. \(40x^2 \times 1 = 40x^2\)

Adding these gives: \(54 - 120 + 40 = -26\)

Thus, the coefficient of \(x^2\) is \(-26\).