Answer
(a) W = -56.6 J
(b) W = 188 J
(c) W = 0
(d) The net work done on the package is 131 J.
Work Step by Step
(a) We can find the work done by friction.
$W = F_f~d~cos(180^{\circ})$
$W = F_N~\mu_k~d~cos(180^{\circ})$
$W = mg~cos(\theta)~\mu_k~d~cos(180^{\circ})$
$W = (12.0~kg)(9.80~m/s^2)~cos(53.0^{\circ})(0.40)(2.00~m)~cos(180^{\circ})$
$W = -56.6~J$
(b) We can find the work done by gravity.
$W = mg~d~cos(37.0^{\circ})$
$W = (12.0~kg)(9.80~m/s^2)(2.00~m)~cos(37.0^{\circ})$
$W = 188~J$
(c) The normal does zero work because the normal force makes a $90^{\circ}$ angle with the direction of motion.
(d) The net work done on the package is $188~J - 56.6~J$, which is $131~J$.
