Answer
$1.02\;\rm m/s$
Work Step by Step
a) First, we need to find the spring constant.
So we need to use the conservation of energy.
$$E_1=E_2$$
$$K_1+U_{s1}=K_2+U_{s2}$$
$$ \color{red}{\bf\not}\frac{1}{2}mv_1^2+ \color{red}{\bf\not}\frac{1}{2}kx_1^2= \color{red}{\bf\not}\frac{1}{2}mv_2^2+ \color{red}{\bf\not}\frac{1}{2}kx_2^2$$
$$ mv_1^2+ kx_1^2= mv_2^2+ kx_2^2$$
Solving for $k$;
$$ mv_1^2-mv_2^2 = kx_2^2-kx_1^2=k(x_2^2- x_1^2)$$
$$k=\dfrac{m(v_1^2-v_2^2) }{x_2^2- x_1^2}$$
Plugging the known;
$$k=\dfrac{(0.300)[(95.4\times 10^{-2})^2-(71.4\times 10^{-2})^2] }{0.06^2- 0.03^2}$$
$$k=\bf 44.48\;\rm N/m$$
Now we can find the maximum speed by applying the total energy law.
$$E_{tot}=K_1+U_{s1}=\frac{1}{2}mv_{\rm max}$$
$$ \color{red}{\bf\not}\frac{1}{2}mv_1^2+ \color{red}{\bf\not}\frac{1}{2}kx_1^2= \color{red}{\bf\not}\frac{1}{2}mv_{\rm max}$$
$$ v_{\rm max}^2=\dfrac{mv_1^2+ kx_1^2}{m}$$
$$ v_{\rm max}=\sqrt{v_1^2+\dfrac{ kx_1^2}{m}}$$
$$ v_{\rm max}=\sqrt{(95.4\times 10^{-2})^2+\dfrac{ (44.48)(0.03)^2}{0.30}}$$
$$ v_{\rm max}=\color{red}{\bf 1.02}\;\rm m/s$$
