Make it with out right steps

Here are the solutions for 5a.i and 5a.ii, presented without explicit step labels.

5a.i. Solve $f(x) = g\left(-\frac{1}{2}\right)$

Given $f(x) = x-2$ and $g(x) = 2x^2 - 1$.

First, evaluate $g\left(-\frac{1}{2}\right)$:

$$g\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^2 - 1$$ $$g\left(-\frac{1}{2}\right) = 2\left(\frac{1}{4}\right) - 1$$ $$g\left(-\frac{1}{2}\right) = \frac{1}{2} - 1$$ $$g\left(-\frac{1}{2}\right) = -\frac{1}{2}$$

Now, set $f(x)$ equal to this value:

$$f(x) = -\frac{1}{2}$$ $$x-2 = -\frac{1}{2}$$ $$x = 2 - \frac{1}{2}$$ $$x = \frac{4}{2} - \frac{1}{2}$$ $$x = \frac{3}{2}$$

The solution is $\boxed{x = \frac{3}{2}}$.

5a.ii. Solve $f(x) + g(x) = 0$

Substitute the expressions for $f(x)$ and $g(x)$:

$$(x-2) + (2x^2 - 1) = 0$$ $$2x^2 + x - 3 = 0$$

Factor the quadratic equation:

$$2x^2 + 3x - 2x - 3 = 0$$ $$x(2x+3) - 1(2x+3) = 0$$ $$(x-1)(2x+3) = 0$$

Set each factor to zero to find the solutions for $x$:

$$x-1 = 0 \implies x = 1$$ $$2x+3 = 0 \implies 2x = -3 \implies x = -\frac{3}{2}$$

The solutions are $\boxed{x = 1 \text{ or } x = -\frac{3}{2}}$.