8) Arrange the three numbers $3, \sqrt{10}, \sqrt{2}+\sqrt{3}$ in ascending order.
(A) $3<\sqrt{10}<\sqrt{2}+\sqrt{3}$
(D) $3<\sqrt{2}+\sqrt{3}<\sqrt{10}$
(B) $\sqrt{2}+\sqrt{3}<3<\sqrt{10}$,
(E) $\sqrt{2}+\sqrt{3}<\sqrt{10}<3$.
(C) $\sqrt{10}<\sqrt{2}+\sqrt{3}<3$
8. The answer is $(\mathbf{D})$.

Recall that if $a>b>0$ then $a^{2}>b^{2}$. Squaring the 3 numbers proposed, we get $9, 10, 5+\sqrt{6}$, from which it immediately follows that $35+2 \sqrt{4}=9$, that is, $\sqrt{2}+\sqrt{3}>3$.

It remains to compare $\sqrt{2}+\sqrt{3}$ with $\sqrt{10}$, that is, $5+2 \sqrt{6}$ with 10, or equivalently, $2 \sqrt{6}$ with 5.

If we compare their squares, we have: $(2 \sqrt{6})^{2}=24<25=5^{2}$, that is, $\sqrt{2}+\sqrt{3}<\sqrt{10}$.

In conclusion, $3<\sqrt{2}+\sqrt{3}<\sqrt{10}$.