Answer
a) $W=(mg)(h_{max}-h)$
b) $W=mgh$
c) $K.E_f=\frac{1}{2}mv_{\circ}^2+mgh$
Work Step by Step
(a) The required work done is given as
$W=Fd$
$W=(mg)(h_{max}-h)$
(b) Work done on ball by gravity when it reaches ground level is given as
$W=Fd$
$W=-(mg)(0-h)$
$W=mgh$
(c) We know that the expression for the final kinetic energy is given as
$K.E_f-K.E-i=W$
$\implies K.E_f=K.E_i+W$
$\implies K.E_f=\frac{1}{2}mv_{\circ}^2+mgh$