= Solution
Write $N=\sum_iI_i$ as in part (ii). For $i\ne j$, the two vertices are both isolated exactly when their combined $2n-3$ incident edges are absent, so
$$
\mathbb E(I_iI_j)=(1-p)^{2n-3}.
$$
Therefore
$$
\boxed{
\mathbb E(N^2)
=n(1-p)^{n-1}
+n(n-1)(1-p)^{2n-3}.
}
$$
Let $A=\mathbb EN=n(1-p)^{n-1}$. The displayed formula gives
$$
\frac{\operatorname{var}(N)}{A^2}
=\frac1A+\frac{n-1}{n(1-p)}-1
=\frac1A+\frac{p-1/n}{1-p}.
$$
Now take $p=c\log n/n$ with $c<1$, and choose $\alpha>1$ such that $\alpha c<1$. For all sufficiently large $n$, the supplied inequality gives $1-p\geq e^{-\alpha p}$, whence
$$
A\geq ne^{-\alpha p(n-1)}
=n^{\,1-\alpha c(n-1)/n}
\longrightarrow\infty.
$$
Also $p\to0$, so both terms in the variance ratio tend to zero. Part (iv) now applies and proves the lower side of the isolated-vertex threshold:
$$
\boxed{\mathbb P(N=0)\longrightarrow0}.
$$
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