= Solution
The supplied bracket gives $0<e_0=1.5-\sqrt2<0.1$. Part (b) ensures $x_n>\sqrt2>1.4$, so each error satisfies
$$
0<e_{n+1}<\frac{e_n^2}{2.8}.
$$
Writing $d_n=e_n/2.8$ gives $d_{n+1}<d_n^2$ and $d_0<0.1/2.8=1/28$. After three steps the exponent has doubled three times:
$$
\boxed{0<e_3<\frac{2.8}{28^8}\approx7.4113\times10^{-12}<10^{-11}.}
$$
This rigorous <error bound> uses only the given bracket, not advance knowledge of the digits of $\sqrt2$.
The actual rational iterates are $x_1=17/12$, $x_2=577/408$, and $x_3=665857/470832\approx1.414213562374690$. They illustrate the bound, but the proof comes from the exact error identity and the positive lower bound on the denominators.
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