Answer
a) Down.
b) $76.5\;\rm kg$
c) $39.2\;\rm m$
Work Step by Step
a)
When Henry stands in the elevator there are only two forces acting on him, his own weight (which is the gravitational force due to Earth's gravitational pull) and the normal force from the elevator ground (which we can call the apparent weight).
1)- This normal force increases (be more than his own weight) when Henry speeds up upward or slows down downward since the net force now is upward and hence the acceleration is upward.
2)- But this normal force decreases if he speeds up downward or slows down upward since the net force now pointing down and his weight is constant so the changeable thing here is the normal force (the apparent weight).
In the given figure, we can see that his weight (apparent weight) increases, so he might be speeding up upward or slowing down downward. To guess which is the right case, let us analyze the given data.
We know that elevators undergo three stages of motion, firstly speeding up (up or down) then moving at constant velocity then slowing down. All elevators have these 3 stages and the second stage is mostly the longest stage
Now we can see that the longest trip of Henry is from $t=2$ s to $t=10$ s which is the second stage of elevator motion (constant velocity). In this stage, the net force exerted on him is zero and hence the normal force exerted on him and his real own eight are equal.
Thus,
$$W_{Henry}=750\;\rm N\tag 1$$
Now we can see that his apparent weight decreases from his real weight for the first two seconds while the elevator is speeding up. This means that the normal force decreased while the elevator is speeding up, therefore, the elevator is speeding up downward as we mentioned in the first case of point 2 above.
b)
We found Henry's weight in (1) and we know that
$$W_{}=mg$$
So his mass is
$$m=\dfrac{W_{Henry}}{g}=\dfrac{750}{9.8}=\color{red}{\bf 76.5}\;\rm kg$$
c)
During the first stage of motion, the net force exerted on Henry is downward. We can choose downward to be our positive direction for now.
$$\sum F_y=F_G- F_{n1}=ma_{y1}$$
Thus,
$$a_{y1}=\dfrac{F_G- F_{n1}}{m}=\dfrac{750-600}{\dfrac{750}{9.8}}=\color{blue}{\bf 1.96}\;\rm m/s^2$$
Thus, the final velocity of this stage is
$$v_{y1}=v_{iy}+a_{y1}t_1=0+1.96\cdot (2-0)=\color{blue}{\bf3.92}\;\rm m/s$$
Thus,
$$\Delta y_1=\overbrace{v_{iy}t_1}^{0}+\frac{1}{2}a_{y1}t_1^2=\frac{1}{2}\cdot 1.96\cdot 2^2=\color{red}{\bf3.92}\;\rm m$$
During the second stage, the acceleration is zero since the elevator moves at a constant speed.
We found Henry's speed in this stage which is the final speed of the first stage.
Thus, the distance traveled during this stage is
$$\Delta y_2=v_{1y}t_2=3.92\cdot (10-2)=\color{red}{\bf 31.36}\;\rm m$$
In the third stage, when the elevator slows down, we need to find its acceleration as well.
The net force now is upward which we chose to be the negative direction.
$$a_{y3}=\dfrac{ F_G -F_{n3} }{m}=\dfrac{750-900 }{\dfrac{750}{9.8}}=\color{blue}{\bf -1.96}\;\rm m/s^2$$
Thus, the distance traveled during this stage is
$$\Delta y_3=v_{y1}t_3+\frac{1}{2}a_{y3}t_3^2$$
$$\Delta y_3=(3.92\cdot 2_- \frac{1}{2}\cdot 1.96\cdot 2^2=\color{red}{\bf 3.92}\;\rm m$$
Recall that the three distances, $\Delta y_1,\; \Delta y_2$, and $\Delta y_3$, are downward.
Thus the total distance traveled by Henry during the 12 seconds is given by
$$\Delta y= \Delta y_1+\Delta y_2+\Delta y_3=3.92+31.36+3.92$$
$$\Delta y=\color{magenta}{\bf 39.2}\;\rm m$$
