Answer
$$2.858 \times 10^{-12}$$
Work Step by Step
Since
\begin{aligned}
f(x) &=\sqrt{1+x} & & f(8)=3 \\
f^{\prime}(x) &=\frac{1}{2 \sqrt{1+x}} & & f^{\prime}(8)=\frac{1}{6} \\
f^{\prime \prime}(x) &=-\frac{1}{4(1+x)^{\frac{3}{2}}} & & f^{\prime \prime}(8) =-\frac{1}{108} \\
f^{\prime \prime \prime}(x) &=\frac{3}{8(1+x)^{\frac{3}{2}}} & & f^{\prime \prime \prime}(8) =\frac{1}{648}
\end{aligned}
Then
$$T_{3}(x)=3+\frac{1}{6}(x-8)-\frac{1}{216}(x-8)^{2}+\frac{1}{3888}(x-8)^{3}$$
Since $$\left|f(x)-T_{n}(x)\right| \leq K \frac{|x-a|^{n+1}}{(n+1) !} $$
\begin{aligned}
\left|\sqrt{9.02}-T_{3}(8.02)\right| &\leq \frac{5}{11664} \frac{|8.02-8|^{3+1}}{(3+1) !} \\
\left|\sqrt{9.02}-T_{3}(8.02)\right| & \leq \frac{5}{11664} \frac{(0.02)^{4}}{(4) !} \\
&=\frac{5}{11664} \frac{1.6 \times 10^{-7}}{(24)} \\
& \approx\left[2.858 \times 10^{-12}\right]
\end{aligned}
To check
\begin{aligned}
\left|\sqrt{9.02}-T_{3}(8.02)\right| & \approx|\sqrt{9.02}-3.003331484| \\
& \approx 2.84 \times 10^{-12}<2.858 \times 10^{-12}
\end{aligned}
