We'll start by explaining the modulus |x+4|.

Case 1: |x+4| = x + 4 for x+4>=0

We'll solve the inequality x+4>=0:

x+4>=0

x >= -4

The interval of admissible values for x is: [-4,+infinite).

Case 2: |x+4| = -x - 4 for x+4<0

x<-4

The interval of admissible values for x is: (- infinite, -4).

Now, we'll solve the equation in both cases:

1) [f(x)]^2 = 4 for x>=-4

f(x) = x + 4 - 3

f(x) = x + 1

(x+1)^2 = 4

We'll expand the square:

x^2 + 2x + 1 - 4 = 0

x^2 + 2x - 3 = 0

We'll apply the quadratic formula:

x1 = [-2+sqrt(4 + 12)]/2

x1 = (-2+4)/2

x1 = 1

x2 = -3

Since both solutions belong to the interval [-4,+infinite), they are accepted.

2) [f(x)]^2 = 4 for x<-4

f(x) = -x - 4 - 3

f(x) = -x - 7

(-x - 7)^2 = 4

We'll expand the square:

x^2 + 14x + 49 - 4 = 0

x^2 + 14x + 45 = 0

We'll apply the quadratic formula:

x1 = [-14+sqrt(196-180)]/2

x1 = (-14+4)/2

x1 = -5

x2 = -9

Since both solutions belong to the interval (- infinite, -4), they are accepted.

The equation [f(x)]^2 = 4 has the solutions: {-9;-5;-3;1}