(b) There's a vertical component of force due to the boy pulling the box. This plus the normal force exerted by the ground, F_N, will equal the weight (mg). If you draw a free body diagram, you'd have the weight acting down, and the normal force and the vertical component of the 30N force acting up.

mg = F_N + 30*sin30      (where 30*sin30 is the vertical component of the force exerted by the boy)

F_N = mg - 30*sin30

(a) This can be found by calculating the horizontal component of the force applied and then dividing that by the mass of the box. As you may have noticed, horizontal components usually are found by using cos. it's no different here.

F_hor=30N*cos(30º)

F_hor=26N

Now that we have F_hor we divide it by the mass of the box to get acceleration:

a=F_hor/m

a=26N/28kg

a=0.93m/s²

(b) The normal force, as we recall, is equal and opposite of the force of gravity. F_g=mg; m is the mass of the box and g is the acceleration due to gravity so we have:

F_g=28kg*9.81m/s²

F_g=274.9N

So since F_N=-F_g we have that F_N=-274.9N. 

But we also need to take into account the veritcal component of the pull on the box (this force counteracts F_N). To do that we find the vertical component through the Soa part of SoaCahToa:

sinΘ=o/h

sin(30º)=F_ver/30N

30N*sin(30º)=F_ver

15N=F_ver

Now we just add F_ver and F_N to get the real F_N.

Enjoy! =]