Solution

Following the author, we denote total energy by $E$ and kinetic energy by $\mathcal{E}$.

At the threshold, all produced particles are at rest relative to each other, i.e., in the center-of-mass frame their total energy is minimal and equal to the sum of their rest masses, and the momentum is zero. We find the threshold energies for the three cases; naturally, all energies not less than the threshold are possible.

$\textbf{1.}$
Such problems are typically solved as follows: write the invariant for the whole system before and after the pair production. Then, to avoid solving complicated equations for the velocity of the fast particle, it is convenient to write the invariant separately for the fast particle to express its momentum.
$$I_1 = \frac{(E_1 + m_p c^2)^2}{c^2} - p_1^2 = (2m_p c + m_{\pi^0} c)^2 \tag{1}$$
$$(m_p c)^2 = \frac{E_1^2}{c^2} - p_1^2 \tag{2}$$
$$E_1 = \mathcal{E}_1 + m_p c^2 \tag{3}$$
Solving this system:
$$\frac{(E_1 + m_p c^2)^2}{c^2} - \frac{E_1^2}{c^2} = (2m_p c + m_{\pi^0} c)^2 - (m_p c)^2$$
$$2E_1 m_p c^2 = c^4 \left( (2m_p + m_{\pi^0})^2 - 2m_p^2 \right)$$
$$E_1 = c^2 \frac{(2m_p + m_{\pi^0})^2 - 2m_p^2}{2m_p}$$
$$\mathcal{E}_1 = E_1 - m_p c^2 = c^2 \frac{(2m_p + m_{\pi^0})^2 - 4m_p^2}{2m_p} = c^2 \frac{(2m_p + m_{\pi^0})^2 - 4m_p^2}{2m_p}$$
$$\mathcal{E}_1 = m_{\pi^0} c^2 \frac{m_{\pi^0} + 4m_p}{2m_p} \tag{4}$$

$\textbf{2.}$
Cases 1 and 2 differ only in the mass of the mesons, so the answer will be
$$\mathcal{E}_2 = m_{\psi} c^2 \frac{m_{\psi} + 4m_p}{2m_p} \tag{5}$$

$\textbf{3.}$
Here it is completely analogous:
$$I_3 = \frac{(E_3 + m_p c^2)^2}{c^2} - p_3^2 = (4m_p c)^2 \tag{6}$$
$$(m_p c)^2 = \frac{E_3^2}{c^2} - p_3^2 \tag{7}$$
Substituting (7) into (6):
$$\frac{(E_3 + m_p c^2)^2}{c^2} - \frac{E_3^2}{c^2} = (4m_p c)^2 - (m_p c)^2$$
$$2E_3 m_p = c^2 \left( (4m_p)^2 - 2m_p^2 \right)$$
$$E_3 = 7 m_p c^2 \quad \Rightarrow \quad \mathcal{E}_3 = 6 m_p c^2$$

Note: Savchenko has a typo in the answer.