1. Let $\lambda=\{x \mid x=a+b \sqrt{3}, a, b \in \mathbf{Z}\}$, if $x=7+a \sqrt{3} \in \lambda$, and $\frac{1}{x} \in \lambda$, then the value of $a$ is
$-、 1 . a$ 的值为 $\pm 4$.

The value of $a$ is $\pm 4$.

$x=7+a \sqrt{3} \in A, \therefore \quad a$ 是整数.

$x=7+a \sqrt{3} \in A, \therefore \quad a$ is an integer.

又 $\frac{1}{x}=\frac{1}{7+a \sqrt{3}}=\frac{7}{49-3 a^{2}}-\frac{a}{49-3 a^{2}} \sqrt{3}$, 要 $\frac{1}{x} \in A$,必须 $\frac{7}{49-3 a^{2}}, \frac{a}{49-a^{2}}$ 都是整数,只能 $49-$ $3 a^{2}=1$, 由此得 $a= \pm 4$.

Also, $\frac{1}{x}=\frac{1}{7+a \sqrt{3}}=\frac{7}{49-3 a^{2}}-\frac{a}{49-3 a^{2}} \sqrt{3}$. For $\frac{1}{x} \in A$, it is necessary that $\frac{7}{49-3 a^{2}}, \frac{a}{49-a^{2}}$ are both integers, which can only be true if $49-3 a^{2}=1$, leading to $a= \pm 4$.