(a)

Since the energy of decayed particle is \(M\), energy conservation gives \[E_1 + E_2 = M. \tag{1}\] Subtracting the sides of the equations \(E_1^2=p^2+m_1^2,\ E_2^2=p^2+m_2^2\), we have \[(E_1-E_2)(E_1+E_2) = m_1^2-m_2^2.\] From the equation (1), \[E_1-E_2 = \frac{m_1^2-m_2^2}{M}. \tag{2}\] Adding the sides of the equations \(E_1^2=p^2+m_1^2,\ E_2^2=p^2+m_2^2\), we have \[\begin{split} p^2 &= \frac{1}{2}\left[E_1^2+E_2^2-m_1^2-m_2^2\right] \\ &= \frac{1}{4}\left[(E_1+E_2)^2+(E_1-E_2)^2-2(m_1^2+m_2^2)\right]. \end{split}\] From the equations (1) and (2), \[\begin{split} p^2 &= \frac{1}{4}\left[M^2+\frac{(m_1^2-m_2^2)^2}{M^2}-2(m_1^2+m_2^2)\right] \\ &= \frac{1}{4M^2}\left[M^4-2M^2(m_1^2+m_2^2)+(m_1^2-m_2^2)^2\right]. \end{split} \tag{3}\]

(b)

Take the limit \(m_2\to0\) in the equation (3), we have \[\begin{split} p^2 &= \frac{1}{4M^2}\left[M^4-2M^2 m_1^2+m_1^4\right] \\ &= \frac{1}{4M^2}\left[M^2-m_1^2\right]^2 \\ \end{split}\] Here, since \(M>E_1>m_1\) holds, we have \[p = \frac{M^2-m_1^2}{2M}.\]

(c)

Adding the sides of the equations (1) and (2), we have \[\begin{split} E_1 &= \frac{1}{2}\left(M + \frac{m_1^2-m_2^2}{M}\right) \\ &= \frac{M^2+m_1^2-m_2^2}{2M}. \end{split}\] Subtracting the sides of the equations (1) and (2), we have \[\begin{split} E_2 &= \frac{1}{2}\left(M - \frac{m_1^2-m_2^2}{M}\right) \\ &= \frac{M^2-m_1^2+m_2^2}{2M}. \end{split}\]