Step 1: Note the interval restriction.
We are told $\alpha,\beta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, where cosine is always positive. This lets us pick positive square roots.
Step 2: Solve for cos alpha.
From $\cos^4\alpha=\frac{1}{16}$ we take fourth roots: $\cos\alpha = \left(\frac{1}{16}\right)^{1/4}=\frac12$ (positive on this interval).
Step 3: Solve for sin beta.
Similarly $\sin^4\beta=\frac{1}{16}$ gives $|\sin\beta|=\frac12$. We just need $\cos\beta$, which is positive here.
Step 4: Get cos beta from the Pythagorean identity.
$\cos\beta=\sqrt{1-\sin^2\beta}=\sqrt{1-\frac14}=\frac{\sqrt3}{2}$.
Step 5: Add the two cosines.
$\cos\alpha+\cos\beta = \frac12+\frac{\sqrt3}{2} = \frac{1+\sqrt3}{2}$.
Step 6: Recognize the closed form.
Since $\cos 15^\circ=\frac{\sqrt6+\sqrt2}{4}$, we have $\sqrt2\cos 15^\circ=\frac{\sqrt{12}+2}{4}=\frac{2\sqrt3+2}{4}=\frac{\sqrt3+1}{2}$, matching our sum exactly.
\[ \boxed{\sqrt2\cos 15^\circ} \]