Answer
(a) $F_T = \frac{Mg}{2}$
(b) $F_{T1} = \frac{Mg}{2}$
$F_{T2} = \frac{Mg}{2}$
$F_{T3} = \frac{3Mg}{2}$
$F_{T4} = Mg$
Work Step by Step
(a) Since the tension in the rope is the same all along the rope, $F = F_{T1} = F_{T2}$. The tension in $F_{T1}$ and $F_{T2}$ pulls up the piano, therefore the sum of these two tension forces must be at least equal to the weight of the piano:
$F_{T1} + F_{T2} = Mg$
$2 F_T = Mg$
$F_T = \frac{Mg}{2}$
(b) $F_{T1} = F_{T2} = F_T = \frac{Mg}{2}$
$F_{T3} = F_{T1} + F_{T2} + F_T = \frac{3Mg}{2}$
$F_{T4} = F_{T1} + F_{T2} = Mg$