Answer
(a) F = 99.2 N
(b) W = 387 J
(c) W = -387 J
(d) These forces (gravity and the normal force) do zero work.
(e) The total work done on the crate is zero.
Work Step by Step
(a) Since the velocity is constant, the horizontal component of the worker's force $F$ must be equal in magnitude to the force of kinetic friction.
$F~cos(\theta) = F_f$
$F~cos(\theta) = F_N~\mu_k$
$F~cos(\theta) = mg~\mu_k+F~sin(\theta)~\mu_k$
$F~cos(\theta) - F~sin(\theta)~\mu_k = mg~\mu_k$
$F = \frac{mg~\mu_k}{cos(\theta) - sin(\theta)~\mu_k}$
$F = \frac{(30.0~kg)(9.80~m/s^2)(0.25)}{cos(30^{\circ}) - sin(30^{\circ})(0.25)}$
$F = 99.2~N$
(b) $W = F\cdot d~cos(30^{\circ}) = (99.2~N)(4.5~m)~cos(30^{\circ})$
$W = 387~J$
(c) $W = -387~J$
Note that the force of kinetic friction is equal in magnitude to the horizontal component of the applied force, but it opposes the direction of motion.
(d) The normal force and gravity act at a $90^{\circ}$ angle to the direction of motion. Therefore the work these forces do is zero.
(e) The total work done on the crate is zero. Note that the kinetic energy of the crate does not change.