2.013. $\left(\frac{1}{\sqrt{a}+\sqrt{a+1}}+\frac{1}{\sqrt{a}-\sqrt{a-1}}\right):\left(1+\sqrt{\frac{a+1}{a-1}}\right)$.
Solution.

Domain of definition: $\left\{\begin{array}{l}a \geq 0, \\ \sqrt{a}-\sqrt{a-1} \neq 0, \Leftrightarrow\left\{\begin{array}{l}a \geq 0, \\ a>1\end{array}\right. \\ \frac{a+1}{a-1} \geq 0\end{array} \Leftrightarrow a>1\right.$.

Let $X$ be the expression in the first parentheses, $Y$ be the expression in the second parentheses:

$$
\begin{aligned}
& X=\frac{1}{\sqrt{a}+\sqrt{a+1}}+\frac{1}{\sqrt{a}-\sqrt{a-1}}=\frac{\sqrt{a}-\sqrt{a-1}+\sqrt{a}+\sqrt{a+1}}{(\sqrt{a}+\sqrt{a+1})(\sqrt{a}-\sqrt{a-1})}= \\
& =\frac{2 \sqrt{a}+\sqrt{a+1}-\sqrt{a-1}}{a+\sqrt{a(a+1)}-\sqrt{a(a-1)}-\sqrt{(a-1)(a+1)}} \\
& Y=1+\frac{\sqrt{a+1}}{\sqrt{a-1}}=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a-1}} \\
& \text { Then } X: Y=\frac{2 \sqrt{a}+\sqrt{a+1}-\sqrt{a-1}}{a+\sqrt{a(a+1)}-\sqrt{a(a-1)}-\sqrt{(a-1)(a+1)}}: \frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a-1}}= \\
& =\frac{\sqrt{a-1}(2 \sqrt{a}+\sqrt{a+1}-\sqrt{a-1})}{a \sqrt{a+1}+a \sqrt{a}+\sqrt{a}-a \sqrt{a-1}-\sqrt{a-1}+a \sqrt{a-1}-a \sqrt{a}+\sqrt{a}-a \sqrt{a+1}+\sqrt{a+1}}= \\
& =\frac{\sqrt{a-1}(2 \sqrt{a}+\sqrt{a+1}-\sqrt{a-1})}{2 \sqrt{a}+\sqrt{a+1}-\sqrt{a-1}}=\sqrt{a-1} \text {. }
\end{aligned}
$$

Answer: $\sqrt{a-1}$.