Two functions g and h are defined on the set R, of real numbers by g: x x2 and h: x (1)/(x+2), x != -2. Find i) h⁻¹, ii) g o h(-(1)/(2)).

The function $h(x)$ is given by $h(x) = \frac{1}{x+2}$.
To find the inverse, let $y = h(x)$, so $y = \frac{1}{x+2}$.
Swap $x$ and $y$:

$$x = \frac{1}{y+2}$$

Now, solve for $y$:

$$x(y+2) = 1$$ $$xy + 2x = 1$$ $$xy = 1 - 2x$$ $$y = \frac{1 - 2x}{x}$$

Thus, the inverse function is:

$$h^{-1}(x) = \boxed{\frac{1 - 2x}{x}, \quad x \neq 0}$$

The function $h(x)$ is $h(x) = \frac{1}{x+2}$.
Substitute $x = -\frac{1}{2}$ into $h(x)$:

$$h\left(-\frac{1}{2}\right) = \frac{1}{-\frac{1}{2} + 2}$$

To simplify the denominator, find a common denominator:

$$-\frac{1}{2} + 2 = -\frac{1}{2} + \frac{4}{2} = \frac{-1+4}{2} = \frac{3}{2}$$

So,

$$h\left(-\frac{1}{2}\right) = \frac{1}{\frac{3}{2}}$$ $$h\left(-\frac{1}{2}\right) = \frac{2}{3}$$

We need to find $g\left(h\left(-\frac{1}{2}\right)\right)$. From Step 2, we know $h\left(-\frac{1}{2}\right) = \frac{2}{3}$.
The function $g(x)$ is given by $g(x) = x^2$.
Substitute $x = \frac{2}{3}$ into $g(x)$:

$$g\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^2$$ $$g\left(\frac{2}{3}\right) = \frac{2^2}{3^2}$$ $$g\left(\frac{2}{3}\right) = \frac{4}{9}$$

Therefore,

$$g \circ h \left(-\frac{1}{2}\right) = \boxed{\frac{4}{9}}$$