= Solution
Composing the <permutation cycles> from right to left gives
$$
(13)(2457)(815)=(1\,7\,2\,4\,5\,8\,3),
$$
with $6$ fixed. It therefore has \b[order $7$] and, as a seven-cycle, \b[sign $(-1)^6=+1$].
An element of $S_6$ has order $6$ precisely for cycle type $(6)$ or $(3,2,1)$. These contribute
$$
\frac{6!}{6}=120,\qquad
\binom63(3-1)!\binom32=120,
$$
so there are \b[$240$ elements of order $6$]. Order $3$ has cycle type $(3,1,1,1)$ or $(3,3)$, contributing
$$
\binom63(3-1)!=40,\qquad \frac{6!}{3^2\,2!}=40,
$$
so there are \b[$80$ elements of order $3$].
Inspecting the partitions of $9$ shows that the largest cycle-length least common multiple available in $S_9$ is $20$, from the partition $5+4$, but such a permutation is odd. Among integers from $16$ through $20$, orders $17$ and $19$ would require cycles of those prime lengths, order $16$ requires a cycle length divisible by $16$, and order $18$ requires disjoint cycles using at least $9+2>9$ letters. The disjoint product of a five-cycle and a three-cycle is even and has order $15$. Hence \b[the greatest order in $A_9$ is $15$].
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