An alternative construction of such a group: start with the free product of two copies of $A_5$: $G=A_5*A_5$. This group is virtually free, so any torsion-free subgroup is free. Consider the kernel of any group homomorphism from $G$ to $A_5$ that is an isomorphism when restricted to each of the two copies of $A_5$. This kernel is free of rank 59, and since $G$ does contain copies of $A_5$ that map isomorphically to the quotient $A_5$, $G$ is isomorphic to $F_{59}\rtimes A_5$.

An alternative construction of such a group: let $G$ be the free product of two copies of $A_5$, i.e., $G=A_5*A_5$. This group $G$ is perfect and it is also virtually free, so that any torsion-free subgroup is free. Consider the kernel of any group homomorphism from $G$ to $A_5$ that is an isomorphism when restricted to each of the two copies of $A_5$. This kernel is free of rank 59, and since $G$ does contain copies of $A_5$ that map isomorphically to the quotient $A_5$, $G$ is isomorphic to $F_{59}\rtimes A_5$.

The calculation that the rank of the kernel is 59 is as follows: the Bass-Serre tree for $G=A_5*A_5$ is a tree with $G$-action with one free orbit of edges, and two orbits of vertices, each with stabilizer a copy of $A_5$. Since the kernel of the map $G\rightarrow A_5$ has trivial intersection with the vertex stabilizers, it acts freely on the tree. It thus has two free orbits of vertices and 60 free orbits of edges. So the quotient graph is two vertices joined by 60 edges, which has fundamental group $F_{59}$.