Answer
(a) $F = (m_A+m_B)~g~\mu_k$
(b) $T = m_A~g~\mu_k$
Work Step by Step
(a) Since the velocity is constant, the force of the pull $F$ must be equal in magnitude to the total force of kinetic friction acting on both blocks.
$F = m_A~g~\mu_k+m_B~g~\mu_k$
$F = (m_A+m_B)~g~\mu_k$
(b) Since the velocity is constant, the force of the tension $T$ must be equal in magnitude to the force of kinetic friction acting on block A.
$T = m_A~g~\mu_k$
