To analyze the solution sets of the given inequalities, we need to check the discriminant $\Delta$ and the coefficient of $x^2$ to determine the nature of the roots and consequently, the solution set of each inequality.

### Option A: $x^{2}+6x+10 \gt 0$

1. Calculate the discriminant: $\Delta = b^{2}-4ac = 6^{2}-4\times 1\times 10 = 36 - 40 = -4 \lt 0$.
2. Since $\Delta \lt 0$ and the coefficient of $x^2$ (which is $1$) is positive, the parabola opens upwards and does not intersect the x-axis. Therefore, the inequality holds for all real numbers.

$\boxed{A \text{ is correct}}$.

### Option B: ${x}^{2}+2\sqrt{5}x+5 \gt 0$

1. Calculate the discriminant: $\Delta = (2\sqrt{5})^{2}-4\times 1\times 5 = 20 - 20 = 0$.
2. Since $\Delta = 0$, the equation has a double root: ${x}_{1}={x}_{2}=-\sqrt{5}$.
3. The parabola opens upwards (coefficient of $x^2$ is positive), touching the x-axis at one point. Hence, it is positive for all $x$ except at $x=-\sqrt{5}$.

$\boxed{B \text{ is incorrect}}$.

### Option C: $-x^{2}+x-2 \lt 0$

1. Calculate the discriminant: $\Delta = 1^{2}-4\times (-1)\times (-2) = 1 - 8 = -7 \lt 0$.
2. Since $\Delta \lt 0$ and the coefficient of $x^2$ (which is $-1$) is negative, the parabola opens downwards and does not intersect the x-axis. Therefore, the inequality holds for all real numbers.

$\boxed{C \text{ is correct}}$.

### Option D: $2x^{2}-3x-3 \lt 0$

1. Calculate the discriminant: $\Delta = (-3)^{2}-4\times 2\times (-3) = 9 + 24 = 33 \gt 0$.
2. Since $\Delta \gt 0$, the equation has two distinct real roots: $x=\frac{3 \pm \sqrt{33}}{4}$.
3. As the parabola opens upwards (coefficient of $x^2$ is positive), the inequality holds between the roots.

$\boxed{D \text{ is incorrect}}$.

Therefore, considering the analysis above, the correct options are $\boxed{AC}$.