= Solution
For $R=(3/4)cT$, part (ii) requires $\gamma_Bv_B/c\geq3/4$, so
$$
\frac{v_B}c\geq\frac35,\qquad \gamma_B\geq\frac54.
$$
The speed formula in part (i) can be written $v_B/c=(\mathcal E^2-m_B^2c^4)/(\mathcal E^2+m_B^2c^4)$. Requiring it to be at least $3/5$ gives $\mathcal E\geq2m_Bc^2$, hence $\gamma_A m_A\geq m_B$.
Since $\gamma_A\geq1$ already, the exact minimum is
$$
\boxed{v_{A,\min}=\begin{cases}c\sqrt{1-(m_A/m_B)^2},&m_B>m_A,\\0,&m_B\leq m_A.\end{cases}}
$$
In the second case, the rest energy of the incoming pair already suffices for the required outgoing speed. Keeping this mass-ratio case avoids taking the square root of a negative number.
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