Solution:
We first claim that $\alpha := 1+\cos^{2} x+\sin^{4} x = 1+\sin^{2} x+\cos^{4} x$. Indeed, note that
$$\sin^{4} x - \cos^{4} x = (\sin^{2} x + \cos^{2} x)(\sin^{2} x - \cos^{2} x) = \sin^{2} x - \cos^{2} x$$
which is the desired after adding $1+\cos^{2} x+\cos^{4} x$ to both sides.

Hence, since $\sin \alpha = \frac{13}{14}$, we have $\cos \alpha = \pm \frac{3 \sqrt{3}}{14}$. It remains to determine the sign. Note that $\alpha = t^{2} - t + 2$ where $t = \sin^{2} x$. We have that $t$ is between $0$ and $1$. In this interval, the quantity $t^{2} - t + 2$ is maximized at $t \in \{0,1\}$ and minimized at $t = 1/2$, so $\alpha$ is between $7/4$ and $2$. In particular, $\alpha \in (\pi/2, 3\pi/2)$, so $\cos \alpha$ is negative. It follows that our final answer is $-\frac{3 \sqrt{3}}{14}$.