Solution
a)
Free Body Diagram

Let the small blue point be the box
Two forces act on the box: the weight W of the box and N the force normal to and exerted by the inclined plane on the box (blue point)
b)
Use system of axes x-y as shown to write all forces in their components form

Vectors N, W and a in components form:
     N = (0 , |N|)
     W = (Wx , Wy) = (|W| cos (27°) , - |W| sin (27°))
     a = (a_x , a_y) = (|a| , 0) , box moving down the inclined plane in the direction of positive x hence a_y = 0.
Use Newton's second law to write that the sum of all forces on the box is equal to the mass times the acceleration (vector equation)
     W + N = M a , M is the mass of the box
In components form, the above equation becomes
     (|W| sin (27°) , - |W| cos (27°)) + (0 , |N|) = M (|a| , 0)
For two vectors to be equal, their components must be equal. Hence
     x-components are equal : |W| sin (27°) + 0 = M |a|
     y-components are equal : - |W| cos (27°) + |N|= 0
     M |a| = |W| sin (27°)
weight: |W| = M g ; g = 10 m/²
|a| = M g sin (27°) / M = g sin (27) m/s^2 ≈ 4.5 m/s^2
c)
|N| = |W| cos (27°) = 2 × 10 cos (27°) ≈ 17.8 N

Free Body Diagram
The box is the small blue point. In the diagram below, W is the weight of the box, N the normal force exerted by the inclined plane on the box, F_a is the force applied to have the box in equilibrium and F_s the force of friction opposite F_a.

b)

The box is at rest, hence its acceleration is equal to 0, therefore the sum of all forces acting on the box is equal to its mass times its acceleration which is zero. ( Newton's second law)
     F_a + W + N + F_s = 0
The components form of all forces (vectors) acting on the box are:
     F_a = (30 , 0)
     |W| = 5 × 10 = 50 N
     W = (W_x , W_y) = ( - |W| sin(30°) , - |W| cos(30°)) = (- 50 sin (30°) , - 50 cos (30°) )
     N = (0 , |N|)
     F_s = (-|F| , 0)
The F_a + W + N + F_s = 0 in components form:
     (30 , 0) + (- 50 sin (30°) , - 50 cos (30°) ) + (0 , |N|) + (-|F| , 0) = 0
     x components: 30 - 50 sin(30) + 0 - |F| = 0
     |F| = - 50 sin(30) + 30 = 5 N
c)
y components equation:
     0 - 50 cos (30) + |N| + 0 = 0
     |N| = 50 cos (30) = 25 √3 ≈ 43.3 N

Free Body Diagram
T tension of string, W weight of the box, N force normal to and exerted by the inclined plane on the box, F_s is the force of friction

Equilibrium: W + T + N + F_s = 0
Forces represented by their components
     W = (Wx , Wy) = ( - M g sin(35°) , - M g cos(35°))
     T = (Tx , Ty) = (|T| cos (25°) , |T| sin (25°) )
     N = (0 , Ny) = (0 , |N|)
F_s = (- |F_s| , 0) = ( - μ_s |N| , 0) , where μ_s is the coefficient of friction between the box and the inclined plane.
Sum of x components = 0
- M g sin(35°) + |T| cos (25°) + 0 - μ_s |N| = 0
which may be rewritten as
     |T| cos (25°) = μ_s |N| + M g sin(35°)
sum of y components = 0
     - M g cos(35°) + |T| sin (25°) + |N| + 0 = 0
     |T| sin (25°) = M g cos(35°) - |N|
We now need to solve the system of two equations with two unknowns |T| and |N|.
     |T| cos (25°) = μ_s |N| + M g sin(35°)         (equation 1)
     |T| sin (25°) = M g cos(35°) - |N|         (equation 2)
solve equation 2 above for |N| to get
     |N| = M g cos(35°) - |T| sin (25°)
Substitute |N| by M g cos(35°) - |T| sin (25°) in eq 1 to get
     |T| cos (25°) = μ_s [ M g cos(35°) - |T| sin (25°) ] + M g sin(35°)
rewrite above equation as follows
     |T| [ cos (25°) + μ_s sin (25°) ] = μ_s M g cos(35°) + M g sin(35°)
Solve for |T|

Free Body Diagram
forces acting on the box (purple point) and their components.
W weight of the box, N the normal force exerted by the inclined plane on the box, F_a the applied force so that the box is lowered at constant speed and F_k kinetic force of friction (box is moving downward, force of friction opposite motion).

Components of all forces
     N = (0 , Ny) = (0 , |N|)
     F_a = (|F_a| , 0)
     F_k = (|F_k| , 0)
     W = (- M g sin α , - M g cos α )
Second Newton's law: constant speed means acceleration = 0, sum of all forces equals mass times acceleration, hence
     W + N + F_a + F_k = 0 (vector form)
component equations
     x components: 0 + |F_a| + |F_k| - M g sin α = 0         (eq 1)
     y components: |N|+0 + 0 - M g cos α = 0         (eq 2)
equation (2) gives
     |N| = M g cos α
kinetic force of friction formula: |Fk| = μ_k |N| = μ_k M g cos α
substitute |Fk| by μ_k M g cos α into equation (1)
     |Fa| = M g sin α - |Fk| = M g sin α - μ_k M g cos α
M = 100 Kg, g = 10 m/s^2
sin α = 2/4 = 1/2
sin α = √(1 - sin² α) = √3 / 2
     |Fa| = 1000 (1/2 - 0.45 √3 / 2) ≈ 110.3 N

     W = (-M g sin α , -M g cos α )
     N = (0 , |N|)
     F_a = (|F_a| cos α , - |F_a| sin α)
     F_s = (|F_s| , 0) = (μ_s|N| , 0)
sum of all x components = 0 gives:
     - M g sin α + 0 + |F_a| cos α + μ_s|N| = 0         (equation 1)
sum of all ycomponents = 0 gives:
     - M g cos α + |N| - |F_a| sin α + 0 = 0         (equation 2)
Equation (2) gives:
     |N| = |F_a| sin α + M g cos α
Substitute |N| by |F_a| sin α + M g cos α in equation (1) to get
     - M g sin α + |F_a| cos α + μ_s( |F_a| sin α + M g cos α ) = 0
     |F_a| (cos α + μ_s sin α ) = M g sin α - μ_s M g cos α