SOLUTION:
Part A
The percent uncertainty in the distance is
\begin{align*}
\text{\%\:uncertainty}_{\text{distance}} & =\frac{25\:\text{m}}{42.188\:\text{km}}\times \frac{1\:\text{km}}{1000\:\text{m}}\times 100\% \\
& =0.0593\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
Part B
The uncertainty in time is
\begin{align*}
\text{\%\:uncertainty}_{\text{time}} & =\frac{1\:\text{s}}{9012\:\text{s}}\times 100\% \\
& =0.0111\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
Part C
The average speed is
\begin{align*}
\text{average speed} & =\frac{42.188\:\text{km}}{9012\:\text{s}}\times \frac{1000\:\text{m}}{1\:\text{km}} \\
& = 4.681\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
Part D
The percent uncertainty in the speed is the combination of uncertainties of distance and time.
\begin{align*}
\text{\%\:uncertainty}_{\text{speed}} & =\text{\%\:uncertainty}_{\text{distance}}+\text{\%\:uncertainty}_{\text{time}} \\
& =0.0593\%+0.0111\% \\
& =0.0704\% \\
\end{align*}
Therefore, the uncertainty in the speed is
\begin{align*}
\delta _{speed} & =\frac{0.0704\%}{100\%}\times 4.681\:\text{m/s} \\
& = 0.003\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
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