= Solution
Reduction modulo $p$ maps $\mathrm{SL}_2(\mathbb Z)$ onto $\mathrm{SL}_2(\mathbb F_p)$, of order $p(p^2-1)$. The image of $\Gamma_1(p)$ is the upper unipotent <subgroup> of order $p$. Thus its index in $\mathrm{SL}_2$ is $p^2-1$, and, since $-I\notin\Gamma_1(p)$ for $p\ge5$, its effective index is
$$
d=\frac{p^2-1}{2}.
$$
Every element of $\Gamma_1(p)$ has <trace> congruent to two. An effective elliptic element of order two or three has <trace> zero or $\pm1$ in a lift, so neither is possible for $p\ge5$. Hence $r_2=r_3=0$.
Represent a <modular cusp> by a primitive column $(a,c)$, modulo sign. Its reduction is a nonzero vector in $\mathbb F_p^2$ modulo sign, and the unipotent <subgroup> acts by $(a,c)\mapsto(a+bc,c)$. For $c=0$, the nonzero values of $a$ give $(p-1)/2$ orbits. For $c\ne0$, $a$ varies freely and $c$ modulo sign gives another $(p-1)/2$ orbits. Thus there are $p-1$ <modular cusps>. The reduction classification is sufficient as well as necessary: completing two primitive columns to determinant-one <matrices> and adjusting the second columns by a translation makes congruent columns related by $\Gamma(p)$.
More explicitly, if a determinant-one <matrix> has first column $(a,c)$, conjugating $T^w$ gives
$$
\begin{pmatrix}1-acw&a^2w\\-c^2w&1+acw\end{pmatrix}.
$$
Its least allowable <cusp width> in $\Gamma_1(p)$ is one for $c\equiv0$ and $p$ otherwise. Both types therefore number $(p-1)/2$, with <cusp widths> one and $p$; their <cusp width> sum is $d$. There are no sign-twisted <modular cusp> periods here, because <trace> two cannot be congruent to minus two for these primes. Substitution gives
$$
\boxed{g(X(\Gamma_1(p)))=1+\frac{p^2-1}{24}-\frac{p-1}{2}=\frac{(p-5)(p-7)}{24}.}
$$
These are the <prime Gamma 1 cusp counts and widths>.
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