b
Applying momentum conservation 

\({m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\)

\(0.1 u + m(0) = 0.1 (0) + m(3)\)

\(0.1 u = 3m\)

\(\frac{1}{2}0.1{u^2} = \frac{1}{2}m{\left( 3 \right)^2}\)

Solving we get, \(u = 3\)

\(\begin{gathered}
  \frac{1}{2}k{x^2} = \frac{1}{2}K{\left( {\frac{x}{2}} \right)^2} + \frac{1}{2}\left( {0.1} \right){3^2} \hfill \\
   \Rightarrow \,\,\,\,\frac{3}{4}k{x^2} = 0.9 \hfill \\
   \Rightarrow \,\,\,\,\frac{3}{2} \times \frac{1}{2}k{x^2} = 0.9 \hfill \\
  \therefore \,\,\frac{1}{2}K{x^2} = 0.6\,J\,\left( {total\,initial\,energy\,of\,the\,spring} \right) \hfill \\ 
\end{gathered} \)