$1.7 \quad \frac{a^{-1}-b^{-1}}{a^{-3}+b^{-3}}: \frac{a^{2} b^{2}}{(a+b)^{2}-3 a b}\left(\frac{a^{2}-b^{2}}{a b}\right)^{1}$;

$$
a=1-\sqrt{2}, b=1+\sqrt{2}
$$
1.7 Using formulas (1.7), (1.8), (1.9) and (1.13), we transform the given expression:

$$
\begin{aligned}
& \frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a^{3}}+\frac{1}{b^{3}}} \cdot \frac{a^{2}-a b+b^{2}}{a^{2} b^{2}} \cdot \frac{a b}{(a-b)(a+b)}= \\
& =\frac{b-a}{b^{3}+a^{3}} \cdot \frac{a^{3} b^{3}}{a b} \cdot \frac{a^{2}-a b+b^{2}}{a^{2} b^{2}} \cdot \frac{a b}{(a-b)(a+b)}= \\
& =-\frac{(a-b)\left(a^{2}-a b+b^{2}\right) a b}{(a+b)\left(a^{2}-a b+b^{2}\right)(a-b)(a+b)}=-\frac{a b}{(a+b)^{2}}
\end{aligned}
$$

For $a=1-\sqrt{2}, b=1+\sqrt{2}$, we find

$-\frac{(1-\sqrt{2})(1+\sqrt{2})}{2^{2}}=\frac{1}{4}$.

Answer: $\frac{1}{4}$.