1. Let $\{x\}$ denote the fractional part of the real number $x$. Given $a=(5 \sqrt{2}+7)^{2017}$. Then $a\{a\}=$ $\qquad$
-1.1 .
Let $b=(5 \sqrt{2}-7)^{2017}$.
Then $0<b<1$, and $a b=1$.
Notice that,
$$
a-b=\sum_{k=0}^{1008} 2 C_{2017}^{2 k+1}(5 \sqrt{2})^{2016-2 k} \times 7^{2 k+1} \in \mathbf{Z} \text {. }
$$

Since $a=(a-b)+b(a-b \in \mathbf{Z}, 0<b<1)$, we have
$$
b=\{a\} \Rightarrow a\{a\}=a b=1 .
$$