= 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 $|z|=\sqrt2$. There $|z^4|=4$, $1\le|z^2-1|\le3$, and $\sqrt2-1\le|z-1|\le\sqrt2+1$.
For $a=1/3$, compare with $z^4$:
$$
\left|\frac13(z-1)(z^2-1)+\frac12\right|\le\sqrt2+\frac32<4.
$$
Therefore <Rouché's theorem> gives four <zeros>.
For $a=12$, compare with $-12(z-1)(z^2-1)$:
$$
|z^4-1/2|\le\frac92<12(\sqrt2-1)\le12|z-1||z^2-1|.
$$
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,
$$
|G|\ge(2-\sqrt2)(3-\sqrt2)=8-5\sqrt2>\frac12.
$$
The last strict inequality follows from $3/2>\sqrt2$. Only the simple <zeros> $1,-1$ of $G$ lie inside the disk. Thus, in the requested order,
$$
\boxed{4,\quad3,\quad2.}
$$
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