 Solution

A convenient version of Rouché's theorem is: if $f,g$ are holomorphic functions on a neighbourhood of a closed disk, and $|g|<|f|$ on its boundary, then $f$ and $f+g$ have the same number of zeros, counted with multiplicity, in the disk. In particular neither has a boundary zero under this comparison.

Put $F_a(z)=z^4-a(z-1)(z^2-1)-1/2$ and use the boundary $|z|=\sqrt{2}$. There $|z^4|=4$, $1\le |z^2-1|\le 3$, and $\sqrt{2}-1\le |z-1|\le \sqrt{2}+1$.

For $a=1/3$, compare with $z^4$:Therefore Rouché's theorem gives four zeros.

For $a=12$, compare with $-12(z-1)(z^2-1)$:The comparison polynomial is $-12(z-1{)}^2(z+1)$, which has three zeros in the disk, counting the double zero at $1$.

For $a=5$, take $G(z)=(z^2-1)(z-2)(z-3)$. Expansion gives $F_5=G+1/2$. On the boundary,The last strict inequality follows from $3/2>\sqrt{2}$. Only the simple zeros $1,-1$ of $G$ lie inside the disk. Thus, in the requested order,