= Solution
Use the <Exact Newton error identity for a quadratic> to expose the cancellation without a long quotient expansion. Put $e_k=x_k-x_*$. Since $f(x_*)=0$, direct expansion gives
$$
f(x_k)=2(x_*-\varepsilon)e_k+e_k^2,\qquad
e_{k+1}=e_k-\frac{2(x_*-\varepsilon)e_k+e_k^2}{2(x_*-\varepsilon+e_k)}
=\frac{e_k^2}{2(x_k-\varepsilon)}.
$$
From part (ii), $e_1=\tfrac12\varepsilon^2+O(\varepsilon^3)$ and $x_1-\varepsilon\to1$. It follows that
$$
e_2=\frac{\frac14\varepsilon^4+O(\varepsilon^5)}{2(1+O(\varepsilon^2))}
=\frac18\varepsilon^4+O(\varepsilon^5).
$$
Thus the stronger result is
$$
\boxed{|x_2-x_*|\sim\frac18|\varepsilon|^4,\qquad n_2=4>3.}
$$
The vanishing lower-order terms are a concrete instance of <quadratic convergence>: to leading order, this step squares the previous error.
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