Answer
$\frac{f^{8}(5)}{40,320}$
Work Step by Step
Since, the Taylor's series of $f$ centered at $a$ is
$f(x)=\Sigma_{n=0}^{\infty}\frac{f^{n}(a)(x-a)^{n}}{n!}$
Thus,
$b_{n}=\frac{f^{n}(a)}{n!}$
Now, $b_{8}=\frac{f^{8}(5)}{8!}=\frac{f^{8}(5)}{40,320}$
