Let's rationalize the denominators of the following fractions:

$$
1^{\circ} . \quad \frac{\sqrt{3}+\sqrt{2}}{\sqrt{5-2 \sqrt{6}}}, \quad 2^{\circ} . \quad \frac{\sqrt{3+\sqrt{5}}}{\sqrt[3]{(4 \sqrt{2}-2 \sqrt{10})^{2}}}
$$
$1^{\circ} . \quad \frac{\sqrt{3}+\sqrt{2}}{\sqrt{5-2 \sqrt{6}}}=(\sqrt{3}+\sqrt{2}) \sqrt{5+2 \sqrt{6}}=\sqrt{(5+2 \sqrt{6})^{2}}=5+2 \sqrt{6}$.

$2^{\circ} . \quad \frac{\sqrt{3+\sqrt{5}}}{\sqrt[3]{(4 \sqrt{2}-2 \sqrt{10})^{2}}}=\sqrt[6]{\frac{(3+\sqrt{5})^{3}}{(4 \sqrt{2}-2 \sqrt{10})^{4}}}=\sqrt[6]{\frac{72+32 \sqrt{5}}{(72-32 \sqrt{5})^{2}}}=\sqrt[6]{\frac{9+4 \sqrt{5}}{8(9-4 \sqrt{5})^{2}}}=\sqrt[6]{\frac{(9+4 \sqrt{5})^{3}}{8(81-80)^{2}}}=\frac{\sqrt{2(9+4 \sqrt{5})}}{2}$.

Note. In solving this problem, on page 44 of the 11th grade, in the 6th, 7th, and 8th lines, $(2-\sqrt{5})$ should be written as $(2+\sqrt{5})$.

Number of solutions: 23.