12. The graph of the function $f(x)=x+\frac{1}{x}$ resembles the "tick" mark that teachers draw when grading assignments, so we often refer to $f(x)$ as the "tick function". Its graph is a hyperbola, with asymptotes given by $l_{1}: x=0$ (the $y$-axis) and $l_{2}: y=x$.
(1) Find the coordinates of the vertices of $C$ and its eccentricity;
(2) Find the coordinates of the foci of $C$.
(1) Taking the bisectors of $l_{1}, l_{2}$ as the $x^{\prime}, y^{\prime}$ axes to establish a new coordinate system $x^{\prime} O y^{\prime}$, as shown in the figure. By symmetry, the curve in the new coordinate system is the standard image of a hyperbola.
Notice that $\tan 22.5^{\circ}=\frac{\sin 45^{\circ}}{1+\cos 45^{\circ}}=\sqrt{2}-1$,
then the $\frac{b}{a}=\sqrt{2}-1, e=\sqrt{1+\frac{b^{2}}{a^{2}}}$
$$
=\sqrt{1+(\sqrt{2}-1)^{2}}=\sqrt{4-2 \sqrt{2}} \text {. }
$$

Solving $\left\{\begin{array}{l}y=(\sqrt{2}+1) x, \\ y=x+\frac{1}{x}\end{array} \Rightarrow A\left(\frac{1}{\sqrt[4]{2}}, \frac{\sqrt{2}+1}{\sqrt[4]{2}}\right)\right.$.
Thus, the coordinates of the vertices of $C$ are $A\left(\frac{1}{\sqrt[4]{2}}, \frac{\sqrt{2}+1}{\sqrt[4]{2}}\right), B\left(-\frac{1}{\sqrt[4]{2}},-\frac{\sqrt{2}+1}{\sqrt[4]{2}}\right)$, and the eccentricity $e=\sqrt{4-2 \sqrt{2}}$.
(2) $a=\sqrt{1+(\sqrt{2}+1)^{2}} \cdot \frac{1}{\sqrt[4]{2}}=\frac{\sqrt{4+2 \sqrt{2}}}{\sqrt[4]{2}}=\sqrt{2+2 \sqrt{2}}, c=a e=\frac{2 \sqrt{2}}{\sqrt[4]{2}}=2 \sqrt[4]{2}$.

And $\sin 67.5^{\circ}=\frac{\sqrt{2}+1}{\sqrt{4+2 \sqrt{2}}}, \cos 67.5^{\circ}=\frac{1}{\sqrt{4+2 \sqrt{2}}}$,
so $F_{1}\left(-\frac{\sqrt{2+2 \sqrt{2}}}{\sqrt{2}+1},-\sqrt{2+2 \sqrt{2}}\right), F_{2}\left(\frac{\sqrt{2+2 \sqrt{2}}}{\sqrt{2}+1}, \sqrt{2+2 \sqrt{2}}\right)$.