= Solution
Use the following standard facts: $A_5$ has order $60$ and is a nonabelian <simple group>; the <coset action> on the cosets of an index-$j$ <subgroup> is a transitive <homomorphism> into $S_j$. If $H<A_5$ has index $j>1$, that action is nontrivial. Its <kernel> is a <normal subgroup>, hence is trivial by simplicity. It would therefore embed $A_5$ in $S_j$. For $j=2,3,4$ this is impossible since $j!<60$. \b[There are no <subgroups> of indices 2, 3 or 4.]
An index-5 <subgroup> has order $12$ by <Lagrange's theorem>. Consider its natural action on the five letters. It cannot be transitive, since an orbit of size 5 would divide its order by the <orbit-stabilizer theorem>. If it had no fixed letter, its orbits would have sizes 2 and 3. It would then be a <subgroup> of the <permutations> preserving that partition, namely $S_2\times S_3$. Exactly half of those 12 <permutations> are even, so $(S_2\times S_3)\cap A_5$ has order 6, too small to contain $H$.
Thus $H$ fixes a letter $i$. The stabilizer of $i$ in $A_5$ is the copy of $A_4$ consisting of even <permutations> of the other four letters; it has order 12. Containment and equality of orders force $H$ to equal that stabilizer. Conversely each such stabilizer has index 5. Consequently
$$
\boxed{\text{The index-5 subgroups are precisely }\operatorname{Stab}_{A_5}(i),\quad i=1,\ldots,5.}
$$
These five <subgroups> are distinct: for $i\ne j$, a <three-cycle> on letters other than $i$ can move $j$, so it belongs to the stabilizer of $i$ but not that of $j$.
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