Ii) Obtain the first four terms of the binomial expansion of (1 - (3)/(2)x)⁻² and find the set of values of x for which the expansion is valid.

The first term is 1.

The second term is $ny$:

$$(-2)\left(-\frac{3}{2}x\right) = 3x$$

The third term is $\frac{n(n-1)}{2!}y^2$:

$$\frac{(-2)(-2-1)}{2 \times 1}\left(-\frac{3}{2}x\right)^2 = \frac{(-2)(-3)}{2}\left(\frac{9}{4}x^2\right) = \frac{6}{2}\left(\frac{9}{4}x^2\right) = 3\left(\frac{9}{4}x^2\right) = \frac{27}{4}x^2$$

The fourth term is $\frac{n(n-1)(n-2)}{3!}y^3$:

$$\frac{(-2)(-2-1)(-2-2)}{3 \times 2 \times 1}\left(-\frac{3}{2}x\right)^3 = \frac{(-2)(-3)(-4)}{6}\left(-\frac{27}{8}x^3\right) = \frac{-24}{6}\left(-\frac{27}{8}x^3\right) = -4\left(-\frac{27}{8}x^3\right) = \frac{108}{8}x^3 = \frac{27}{2}x^3$$

So, the first four terms of the expansion are:

$$1 + 3x + \frac{27}{4}x^2 + \frac{27}{2}x^3$$

The binomial expansion $(1+y)^n$ is valid when $|y| < 1$.
In this expansion, $y = -\frac{3}{2}x$.
So, we must have:

$$\left|-\frac{3}{2}x\right| < 1$$ $$\frac{3}{2}|x| < 1$$ $$|x| < \frac{1}{3/2}$$ $$|x| < \frac{2}{3}$$

This inequality can also be written as:

$$-\frac{2}{3} < x < \frac{2}{3}$$

The first four terms of the expansion are $\boxed{1 + 3x + \frac{27}{4}x^2 + \frac{27}{2}x^3}$.
The expansion is valid for $\boxed{|x| < \frac{2}{3}}$ or $\boxed{-\frac{2}{3} < x < \frac{2}{3}}$.