Inequalities

$(x-2)(x+2) < 0$, so the critical values are $\pm 2$. The parabola opens upwards, so it is below the axis between the roots.

$(x+4)(x-2) \geq 0$, with critical values $-4$ and 2. Opening upwards, the curve is on or above the axis outside the roots.

Rearranging gives $3x^2 - 5x - 2 < 0$, which factorises to $(3x+1)(x-2) < 0$. Critical values $-\tfrac13$ and 2; below the axis means between them.

Expand and rearrange: $x^2 - 4x - 5 > 0$, so $(x-5)(x+1) > 0$. Above the axis means outside the critical values $-1$ and 5.

Multiply by $(x+2)^2$, which is always positive, to get $(2x-1)(x+2) < 0$. Critical values $\tfrac12$ and $-2$, and below the axis means between them. Neither endpoint is included, and $x=-2$ would in any case be undefined.

Multiplying by $(x+3)^2$ gives $(x-1)(x+3) \geq 0$, so $x \leq -3$ or $x \geq 1$. But $x = -3$ makes the original denominator zero, so it must be excluded: $x < -3$ or $x \geq 1$.