Points of equilibrium of the spring will be when no force acts on it.

\(k x=\left(m_1+m_2\right) g\)

\(x=\frac{\left(m_1+m_2\right) g}{k}\)

The new equilibrium position which will be the mean position of \(S.H.M.\) will be simply \(\frac{m_2 g}{k}\)

New amplitude will be maximum displacement from \(\frac{m_2 g}{k}\) which is :

\(A=\frac{\left(m_1+m_2\right) g}{k}-\frac{m_2 g}{k}\)

or \(A=\frac{m_1 g}{k}\)

or \(A=\frac{1 \times 10}{12.5}\)

or \(A=\frac{4}{5} \,m\)

\(\therefore A=0.8 \,m\) or \(80 \,cm\)