= Solution
Interpret the second process as an <independent> copy of the unit-rate <Poisson process>. First, the sums are finite almost surely. On $[0,1]$ there are finitely many arrivals, none at $0$, so their inverse-square contributions are finite. On $[1,\infty)$,
$$
\mathbb E\sum_{Z_i\geq1}Z_i^{-2}=\int_1^\infty z^{-2}\,dz=1.
$$
For completeness, this <expectation> identity follows first for nonnegative simple functions from $\mathbb E N(A)=|A|$, and then for all nonnegative measurable functions by <monotone convergence>. Thus no unproved Poisson-integral formula is needed, and the tail sum is finite almost surely.
By (b), merging the two processes produces a rate-$2$ <Poisson process> with arrival times $V_i$. Its rescaled arrivals $2V_i$ form a unit-rate <Poisson process>: for a measurable set $A$, their count is the merged count on $A/2$, with parameter $2|A/2|=|A|$, and disjoint-set independence is preserved. Consequently
$$
X+Y=\sum_iV_i^{-2}=4\sum_i(2V_i)^{-2}\ \stackrel{d}{=}\ 4X.
$$
Therefore \b[$\boxed{c=4}$]. This is the scaling of an <inverse-power sum over Poisson arrivals> with exponent $2$.
The PDF does not explicitly repeat the second process's rate. If that rate were $\rho$ rather than $1$, the same count argument would give $X+Y\stackrel d=(1+\rho)^2X$. The numerical value $4$ uses the intended independent-unit-rate-copy interpretation.
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