= Solution
Condition on the number $M$ of stars. The <independence> assumptions give $\mathbb E[e^{itF_n}\mid M]=q_n(t)^M$, with the inverse-square one-star <characteristic function> $q_n$ from part (a). Averaging over the <Poisson distribution> therefore gives the <compound Poisson distribution> formula directly:
$$
\mathbb Ee^{itF_n}
=\sum_{k=0}^\infty e^{-n}\frac{n^k}{k!}q_n(t)^k
=\exp\{n(q_n(t)-1)\}
=\exp\{-I_n(t)\}.
$$
Since $I_n(t)\to\sqrt{\pi m/2}\,|t|^{1/2}$, the <Lévy continuity theorem> gives exactly the same <weak convergence of random variables> as in part (a):
$$
\boxed{\phi(t)=\exp\!\left(-\sqrt{\frac{\pi m}{2}}\,|t|^{1/2}\right).}
$$
The random number of stars introduces no additional limiting term: its effect is already contained in the exact exponential formula. This is the <symmetric inverse-power Poisson field on a line> at spatial intensity $1/2$ and exponent $p=2$.
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