1. Simplify $\arctan \sqrt{5}-\frac{1}{2} \arcsin \frac{2}{3}=$
$=1 \cdot \frac{\pi}{4}$.
Let $z_{1}=1+\sqrt{5} \mathrm{i}, z_{2}=-\sqrt{5}+2 \mathrm{i}$, then we have $\arg z_{1}+\frac{1}{2} \arg z_{2}=\frac{1}{2} \arg \left(z_{1}^{2} z_{2}\right)$
$$
\begin{array}{l}
=\frac{1}{2} \arg [(-4+2 \sqrt{5} \mathrm{i})(-\sqrt{5}+2 \mathrm{i})] \\
=\frac{1}{2} \arg (-18 \mathrm{i})=\frac{3 \pi}{4} .
\end{array}
$$

Therefore, $\arg z_{1}+\frac{1}{2} \arg z_{2}$
$$
=\arctan \sqrt{5}+\frac{\pi-\arcsin \frac{2}{3}}{2}=\frac{3 \pi}{4} \text {, }
$$

Thus, the original expression $=\frac{\pi}{4}$.