1. Let $p, q \in \mathbf{R}_{+}$, and satisfy
$$
\begin{array}{l}
\log _{9} p=\log _{12} q=\log _{16}(p+q) . \\
\text { Then } \frac{q}{p}=\text {. }
\end{array}
$$
$-1 . \frac{1+\sqrt{5}}{2}$.
By the change of base formula, we get
$$
\begin{array}{l}
\frac{\ln p}{2 \ln 3}=\frac{\ln q}{2 \ln 2+\ln 3}=\frac{\ln (p+q)}{4 \ln 2} . \\
\text { Then } \frac{\ln p-2 \ln q}{2 \ln 3-2(2 \ln 2+\ln 3)}=\frac{\ln (p+q)}{4 \ln 2} \\
\Rightarrow 2 \ln q-\ln p=\ln (p+q) \\
\Rightarrow q^{2}=p(p+q) \\
\Rightarrow \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2} .
\end{array}
$$

Since $p, q>0$, we have $\frac{q}{p}=\frac{1+\sqrt{5}}{2}$.