I have found for $a_0, a_n, b_n$ but as you can see in picture that in first equation the coefficient $b_n$ is removed and is written $2$ in front of coefficient $a_n$. Almost the same for equation two, where $a_0$ and $a_n$ is removed and $2$ is in front of coefficient $b_n$.
Why is this true?
You get that representations by considering the Fourier cosine series of $e^{a|x|}$ over $(-\pi,\pi)$ or the Fourier sine series of $\text{sign}(x)\,e^{a|x|}$ over $(-\pi,\pi)$. The coefficients are computed from:
$$ \int_{0}^{\pi}e^{ax}\cos(nx)\,dx = \frac{a}{a^2+n^2}\left(-1+(-1)^n e^{a\pi}\right), $$ $$ \int_{0}^{\pi}e^{ax}\sin(nx)\,dx = \frac{n}{a^2+n^2}\left(1-(-1)^n e^{a\pi}\right) $$ that are both consequences (take the real or imaginary part) of $$ \int_{0}^{\pi} e^{ax} e^{nix}\,dx = \left.\frac{e^{(a+ni)x}}{a+ni}\right|_{0}^{\pi}.$$