$$\require{cancel}$$

Exercise 3.1.1

\begin{equation*} \begin{aligned} dx = dx' + vdt' \end{aligned} \end{equation*}

dividing by $$dt$$ gives

\begin{equation*} \begin{aligned} u= \frac{dx}{dt} = \frac{dx' + vdt'}{dt} \end{aligned} \end{equation*}

and since $$dt = dt'$$ this simplifies to

\begin{equation*} \begin{aligned} u = u' + v \end{aligned} \end{equation*}

Exercise 3.2.1

Starting from the law in the unprimed frame and using the coordinate transformation we get

\begin{equation*} \begin{aligned} -k((x' + vt') - (x_c' + vt')) = m\ddot{x}' \end{aligned} \end{equation*}

which simplifies to

\begin{equation*} \begin{aligned} -k(x' - x_c') = m\ddot{x}' \end{aligned} \end{equation*}

thus it has the same form and is therefore invariant under the Galilean transformation.

It might be worth pointing out that $$d/dt = d/dt'$$ and since $$x'$$ only differs by a term linear in $$t$$ from $$x$$, their second derivatives with respect to time are the same (hence the right-hand side in the first line of equation in this solution).