1. If $x=\frac{13}{4+\sqrt{3}}$, then $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+3}{x^{3}-7 x^{2}+5 x+15}=$ $\qquad$
$$
\text { II.1. }-5 \text {. }
$$

Since $x=\frac{13}{4+\sqrt{3}}=4-\sqrt{3}$, then $x^{2}-8 x+13=0$.
$$
\begin{array}{l}
\text { Hence } x^{4}-6 x^{3}-2 x^{2}+18 x+3 \\
=\left(x^{2}+2 x+1\right)\left(x^{2}-8 x+13\right)-10=-10, \\
x^{3}-7 x^{2}+5 x+15 \\
=(x+1)\left(x^{2}-8 x+13\right)+2=2 .
\end{array}
$$

Therefore, the original expression $=\frac{-10}{2}=-5$.