3. Let $a$ be the fractional part of $\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}$, and $b$ be the fractional part of $\sqrt{6+3 \sqrt{3}}-\sqrt{6-3 \sqrt{3}}$. Then the value of $\frac{2}{b}-\frac{1}{a}$ is ( ).
(A) $\sqrt{6}+\sqrt{2}-1$
(B) $\sqrt{6}-\sqrt{2}+1$
(C) $\sqrt{6}-\sqrt{2}-1$
(D) $\sqrt{6}+\sqrt{2}+1$
3. B.

Since $\sqrt{3 \pm \sqrt{5}}=\sqrt{\frac{6 \pm 2 \sqrt{5}}{2}}=\frac{\sqrt{5} \pm 1}{\sqrt{2}}$, we have
$$
\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\frac{(\sqrt{5}+1)-(\sqrt{5}-1)}{\sqrt{2}}=\sqrt{2} \text {. }
$$

Therefore, $a=\sqrt{2}-1$.
Also, $\sqrt{6 \pm 3 \sqrt{3}}=\sqrt{\frac{3}{2}}(\sqrt{3} \pm 1)$, so
$$
\sqrt{6+3 \sqrt{3}}-\sqrt{6-3 \sqrt{3}}=\sqrt{6} \text {. }
$$

Therefore, $b=\sqrt{6}-2$.
Thus, $\frac{2}{b}-\frac{1}{a}=\frac{2}{\sqrt{6}-2}-\frac{1}{\sqrt{2}-1}=\sqrt{6}-\sqrt{2}+1$.