1. In the table below, the average of any three adjacent small squares in the upper row is $1(x \neq 0)$, and the square of any four adjacent small squares in the lower row is also 1. Then the value of $\frac{x^{2}+y^{2}+z^{2}-32}{y z+16}$ is $ـ$. $\qquad$
ii. 3 .
From the problem, we know that in the above row, the numbers in every two small squares separated by one are equal, thus we can deduce
$$
\frac{x+7-y}{3}=1 \text {. }
$$

Similarly, for the row below, we get
$$
\begin{array}{l}
\frac{z-x+9.5-1.5}{4}=1 . \\
\begin{array}{l} 
\therefore y=x+4, z=x-4, \text { then } \\
\frac{x^{2}+y^{2}+z^{2}-32}{y z+16}=\frac{x^{2}+\cdot(x+4)^{2}+(x-4)^{2}-32}{(x+4)(x-4)+16} \\
=\frac{3 x^{2}}{x^{2}}=3 .
\end{array}
\end{array}
$$