Answer
a), and b) See the figures below.
c) $4.9\;\rm m/s^2$
Work Step by Step
a) When the conveyer belt runs at a constant speed, the crate is also moving at a constant speed, so the net force exerted on the crate is zero, see the figure below. The two forces exerted on the crate vertically are the normal force (upward) and the gravitational force (downward). There are no horizontal forces exerted on the crate in this case.
b) When the conveyer belt is speeding up, the crate is also speeding up and we assume it is speeding up without skidding on the conveyer's surface, so the force that prevents the crate from skidding is the static friction force. See the second figure below. The two forces exerted on the crate vertically are the normal force (upward) and the gravitational force (downward). And the only force exerted on it horizontally is the static friction force (rightward).
c) We know that the maximum acceleration without skidding occurs at the maximum static friction force between the crate and the belt.
So,
$$\sum F_x=F_{s,max}=ma_{x,max}$$
and we know that the static friction force is given by
$$F_{s,max}=\mu_sF_n $$
So,
$$\mu_sF_n=ma_{x,max}$$
$$a_{x,max}=\dfrac{\mu_sF_n}{m}\tag 1$$
We also know that the net force exerted on the crate vertically is zero, so
$$\sum F_y=F_n-F_G=0$$
Hence,
$$F_n=F_G=mg$$
Plugging into (1);
$$a_{x,max}=\dfrac{\mu_smg}{m}=\mu_sg$$
Plugging the known;
$$a_{x,max}=0.5\cdot 9.8=\color{red}{\bf4.9}\;\rm m/s^2$$
