$1.27 \frac{2-\sqrt{2}-\sqrt{3}}{2+\sqrt{2}-\sqrt{3}}$.
1.27 To get rid of the irrationality in the denominator, we multiply the numerator and the denominator by the expression conjugate to the denominator:

$$
\begin{aligned}
& \frac{(2-\sqrt{2})-\sqrt{3}}{(2+\sqrt{2})-\sqrt{3}} \cdot \frac{(2+\sqrt{2})+\sqrt{3}}{(2+\sqrt{2})+\sqrt{3}}= \\
& =\frac{(2-\sqrt{2})(2+\sqrt{2})-\sqrt{3}(2+\sqrt{2})+\sqrt{3}(2-\sqrt{2})-3}{(2+\sqrt{2})^{2}-3}= \\
& =\frac{-1-2 \sqrt{3}-\sqrt{3} \cdot \sqrt{2}+2 \sqrt{3}-\sqrt{2} \cdot \sqrt{3}}{3+4 \sqrt{2}}=-\frac{1+2 \sqrt{6}}{3+4 \sqrt{2}} .
\end{aligned}
$$

Performing similar actions again, we find

$$
-\frac{1+2 \sqrt{6}}{3+4 \sqrt{2}} \cdot \frac{3-4 \sqrt{2}}{3-4 \sqrt{2}}=\frac{(1+2 \sqrt{6})(3-4 \sqrt{2})}{23}
$$

Answer: $\frac{(1+2 \sqrt{6})(3-4 \sqrt{2})}{23}$.