$1.36 \frac{x^{4}+x^{2}+x \sqrt{2}+2}{x^{2}-x \sqrt{2}+2}-x \sqrt{2}$.
1.36 To factor the numerator, we perform the following transformation:

$$
\begin{aligned}
& x^{4}+1=x^{4}+2 x^{2}+1-2 x^{2}=\left(x^{2}+1\right)^{2}-(\sqrt{2} x)^{2}= \\
& \left(x^{2}+1+\sqrt{2} x\right)\left(x^{2}+1-\sqrt{2} x\right)
\end{aligned}
$$

Then the numerator will be

$$
\begin{aligned}
& \left(x^{2}+\sqrt{2} x+1\right)\left(x^{2}-\sqrt{2} x+1\right)+\left(x^{2}+\sqrt{2} x+1\right)= \\
& =\left(x^{2}+x \sqrt{2}+1\right)\left(x^{2}-x \sqrt{2}+2\right)
\end{aligned}
$$

Thus, we obtain

$$
\begin{aligned}
& \frac{\left(x^{2}+x \sqrt{2}+1\right)\left(x^{2}-x \sqrt{2}+2\right)}{x^{2}-x \sqrt{2}+2}-x \sqrt{2}= \\
& =x^{2}+x \sqrt{2}+1-x \sqrt{2}=x^{2}+1
\end{aligned}
$$

Answer: $x^{2}+1$.