Let us do it very carefully.
You want to solve the inequality
$1.$ It would be nice to get rid of the denominator. If $x\gt 2$, then the denominator is positive. Multiply both sides by $x-2$. We get the equivalent inequality $x^3-4\le x^2-4$, which is equivalent to $x^3\le x^2$. This is never true for $x\gt 2$.
$2.$ If $x\lt 2$, the denominator is negative. So multiplying both sides by $x-2$ reverses the inequality. Thus we are solving
$x^3-4\ge x^2-4$, or equivalently $x^3\ge x^2$. This can be rewritten as $x^2(x-1)\ge 0$. That gives the possibility $x=0$ or $x\ge 1$.
But recall that the calculation for Case $2$ was done under the assumption $x\lt 2$. So the solutions of our inequality are $x=0$ and $1\le x\lt 2$.
The second inequality was handled almost correctly: there is no reversal issue, since $x^2$ is never negative. However, the left-hand side is not defined at $x=0$, and therefore $0$ should not be part of your solution set.
Another way: The only places that the direction of the inequality could change are (1) the places where we have equality and (2) the places where there is a singularity. For your first problem, there is a singularity at $x=2$. Now find all the places where the inequality direction could change, and use suitable "test points." I imagine you know how to handle this method: if detail is needed it can be supplied.