\begin{enumerate}
\item $\left\vert 5-\dfrac{3}{8}\right\vert=\left\vert\dfrac{40}{8}-\dfrac{3}{8}\right\vert=\left\vert\dfrac{37}{8}\right\vert=\dfrac{37}{8}$.

\medskip

\item $\left\vert 1-\dfrac{13}{7}\right\vert=\left\vert \dfrac{7}{7}-\dfrac{13}{7}\right\vert=\left\vert -\dfrac{6}{7}\right\vert=\dfrac{6}{7}$.

\medskip

\item $\left\vert \sqrt{3}-1\right\vert+\left\vert\sqrt{12}-5\right\vert$.

\begin{itemize}
\item $3>1$ donc $\sqrt{3}>\sqrt{1}$, soit $\sqrt{3}-1>0$ ; ainsi, $\left\vert\sqrt{3}-1\right\vert=\sqrt{3}-1$.
\item $12<25$ donc $\sqrt{12}<\sqrt{25}$, soit $\sqrt{12}-5<0$ ; 

ainsi, $\left\vert\sqrt{12}-5\right\vert=-\left(\sqrt{12}-5\right)=5-\sqrt{12}=5-2\sqrt{3}$.
\end{itemize}

Ainsi, $\left\vert \sqrt{3}-1\right\vert+\left\vert\sqrt{12}-5\right\vert=\sqrt{3}-1+5-2\sqrt{3}=\underline{4-\sqrt{3}}$.

\medskip

\item $\left\vert \sqrt{5}-3\right\vert-\left\vert\sqrt{125}-15\right\vert$.

\begin{itemize}
\item $5<9$ donc $\sqrt{5}<\sqrt{9}$, soit $\sqrt{5}-3<0$ ; ainsi, $\left\vert\sqrt{5}-3\right\vert=-\left(\sqrt{5}-3\right)=3-\sqrt{5}$.
\item $125<225$ donc $\sqrt{125}<\sqrt{225}$, soit $\sqrt{125}-15<0$ ; 

ainsi, $\left\vert\sqrt{125}-15\right\vert=-\left(\sqrt{125}-15\right)=15-\sqrt{125}=15-5\sqrt{5}$.
\end{itemize}

Ainsi,  $\left\vert \sqrt{5}-3\right\vert-\left\vert\sqrt{125}-15\right\vert=3-\sqrt{5}-\left(15-5\sqrt{5}\right)=3-\sqrt{5}-15+5\sqrt{5}=\underline{-12+4\sqrt{5}}$.
\end{enumerate}