1. If the set
$$
\left\{x \mid x^{2}-(a+2) x+2-a<0, x \in \mathbf{N}\right\}
$$

contains exactly one element, then the range of the real number $a$ is $\qquad$
$$
\begin{array}{l}
\text { I. } 1 \cdot\left(\frac{1}{2}, \frac{2}{3}\right] . \\
\text { From } x^{2}-(a+2) x+2-a<0(x \in \mathbf{N}) \\
\Rightarrow x^{2}-2 x+2<a(x+1)(x \in \mathbf{N}) . \\
\text { Since } x+1 \geqslant 1 \text {, then } \\
\frac{x^{2}-2 x+2}{x+1}<a(x \in \mathbf{N})
\end{array}
$$

the solution set contains only one element.
Thus, $\frac{(x+1)(x-3)+5}{x+1}<a(x \in \mathbf{N})$ has a solution set containing only one element, which means
$$
(x+1)+\frac{5}{x+1}-4<a(x \in \mathbf{N})
$$

has a solution set containing only one element.
Notice that, $y=t+\frac{5}{t}$ is monotonically decreasing in the interval $(0, \sqrt{5})$ and monotonically increasing in the interval $(\sqrt{5},+\infty)$,
when $t=2$, $y=\frac{9}{2}$;
when $t=3$, $t=\frac{14}{3}$.
Therefore, if $(x+1)+\frac{5}{x+1}-4<a(x \in \mathbf{N})$

has a solution set containing only one element, then
$$
\frac{9}{2}-4<a \leqslant \frac{14}{3}-4 \Rightarrow \frac{1}{2}<a \leqslant \frac{2}{3} \text {. }
$$