1. Nine positive real numbers $a_{1}, a_{2}, \cdots, a_{9}$ form a geometric sequence, and
$$
a_{1}+a_{2}=\frac{3}{4}, a_{3}+a_{4}+a_{5}+a_{6}=15 .
$$

Then $a_{7}+a_{8}+a_{9}=$ . $\qquad$
1. 112.
Let the common ratio be $q$. Then, from the given conditions, we have
$$
\begin{array}{l}
a_{1}(1+q)=\frac{3}{4} \\
a_{1} q^{2}\left(1+q+q^{2}+q^{3}\right)=15
\end{array}
$$

Dividing the above two equations yields $q^{2}\left(1+q^{2}\right)=20$.
Thus, $q=2, a_{1}=\frac{1}{4}$.
Therefore, $a_{7}+a_{8}+a_{9}=a_{1} q^{6}\left(1+q+q^{2}\right)=112$.