= Solution
First prove the concentration statement with its correct scale. Let $U_i=R_i^2$, the $i$th <uniform order statistic>. The number of sample points at most $z$ is $B_z\sim\operatorname{Bin}(n,z)$. Therefore $U_i>z$ <means> $B_z<i$, and $U_i<z$ <means> $B_z\ge i$, up to probability-zero ties. Take $\varepsilon_n=1/\log n$ and $M=\lceil n^{1/10}\rceil$. Applying <Chernoff bounds> at $z=(1\pm\varepsilon_n)^2i/n$ gives, for an absolute $c_0>0$ and sufficiently large $n$,
$$
\Pr\left(R_i\notin\left[(1-\varepsilon_n)\sqrt{i/n},(1+\varepsilon_n)\sqrt{i/n}\right]\right)\le2e^{-c_0\varepsilon_n^2i}.
$$
When the upper endpoint exceeds one its upper-tail event is empty. A <union bound> makes the total failure <probability> for all $M\le i\le n$ at most $2n\exp(-c_0M/(\log n)^2)=o(1)$. On this event,
$$
\sum_{i=M}^n\frac1{R_i}=(1+o(1))\sqrt n\sum_{i=M}^ni^{-1/2}=(2+o(1))n,
$$
where integral comparison gives $\sum_{i=M}^ni^{-1/2}=2\sqrt n+O(\sqrt M)$.
The first <vertex> block has right endpoint $R_1$ and always contains $L_1$, so its initial chord contributes <vertex degree> two. All its other endpoints are those $L_i$ with $i\ge2$ that fall below $R_1$. Conditional on the right endpoints, these events are <independent> with <probabilities> $R_1/R_i$. Thus
$$
d_1(n)=2+\sum_{i=2}^n\mathbf1_{\{L_i\le R_1\}},\qquad \mu(R):=\mathbb E[d_1(n)\mid R]=2+R_1\sum_{i=2}^nR_i^{-1},\qquad\operatorname{Var}(d_1(n)\mid R)\le\mu(R).
$$
The fewer than $M$ early summands contribute at most $M=o(\sqrt n)$ to this conditional <mean>. For the rest, the uniform estimate and <uniform tightness> of $\sqrt nR_1$ give
$$
\frac{\mu(R)}{\sqrt n}-2\sqrt nR_1\xrightarrow{\mathbb P}0.
$$
To control the actual fluctuations without assuming unconditional <independence>, restrict to the concentration event and $\sqrt nR_1\le A$. Then $\mu(R)\le M+2+C A\sqrt n$. Conditional <Chebyshev inequality> bounds the <probability> of a deviation exceeding $\eta\sqrt n$ by $(M+2+C A\sqrt n)/(\eta^2n)=o(1)$. The discarded endpoint event has limiting <probability> $e^{-A^2}$; let $A\to\infty$. Therefore
$$
\frac{d_1(n)}{\sqrt n}-2\sqrt nR_1\xrightarrow{\mathbb P}0,\qquad\frac{d_1(n)}{\sqrt n}\Rightarrow2W.
$$
Since $W$ has a continuous distribution, the corrected <first-vertex degree in the LCD model> tail is
$$
\boxed{\Pr(d_1(n)\ge y\sqrt n)\longrightarrow e^{-y^2/4}\quad(y>0).}
$$
This disproves the printed exponent $-y^2/8$ for $G_n^{(1)}$. The hint's factor is independently contradicted at $i=n$: its proposed upper bound tends to $1/\sqrt2$, whereas
$$
\Pr\left(R_n\le\frac{1+\varepsilon_n}{\sqrt2}\right)=\left(\frac{(1+\varepsilon_n)^2}{2}\right)^n\longrightarrow0.
$$
Thus the printed concentration event has <probability> tending to zero, not one. Replacing its $2n$ by $n$ leads to the proven exponent $-y^2/4$. The exponent $-y^2/8$ would instead arise for $d_1(2n)/\sqrt n$, but then the <graph> has $2n$ <vertices> and the part (b) endpoint limit at $x/\sqrt n$ becomes $e^{-2x^2}$. It cannot repair all the printed claims with one convention. \b[The literal part (c) is false for the named model; the full corrected limit and the source of the discrepancy are established above.]
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