1) 12.5 N east
There are two forces acting on the box along the horizontal direction:
- The applied force of 15 N east, we indicate it with F
- The force of friction of 2.5 N west, we indicate it with 
Taking east as positive direction, we can write the two forces has

Therefore, the net force on the box will be:

And the positive sign means the direction is east.
2) 
We can solve this part by using Newton's second law:

where
is the net force on the box
m is its mass
a is the acceleration
For the box in this problem,
(east)
m = 5.0 kg
Solving for a, we find the acceleration:

And the direction is the same as the net force (east)