Let's solve the given equations step by step.

### First Equation: \((x+4)^2 - (x-5)^2 = 5(x-4)\)

1. Expand the squares:
\[
(x+4)^2 = x^2 + 8x + 16
\]
\[
(x-5)^2 = x^2 - 10x + 25
\]

2. Substitute these into the equation:
\[
(x^2 + 8x + 16) - (x^2 - 10x + 25) = 5(x-4)
\]

3. Simplify the left-hand side:
\[
x^2 + 8x + 16 - x^2 + 10x - 25 = 5(x-4)
\]
\[
18x - 9 = 5(x-4)
\]

4. Distribute the 5 on the right-hand side:
\[
18x - 9 = 5x - 20
\]

5. Move all terms involving \(x\) to one side and constants to the other:
\[
18x - 5x = -20 + 9
\]
\[
13x = -11
\]

6. Solve for \(x\):
\[
x = -\frac{11}{13}
\]

### Second Equation: \(x(3-x)^2 - 1 = (x+2)^3\)

1. Expand both sides:
\[
(3-x)^2 = 9 - 6x + x^2
\]
\[
x(9 - 6x + x^2) - 1 = (x+2)^3
\]

2. Expand the right-hand side:
\[
(x+2)^3 = x^3 + 6x^2 + 12x + 8
\]

3. Substitute and expand the left-hand side:
\[
x(9 - 6x + x^2) - 1 = x^3 - 6x^2 + 9x - 1
\]

4. Set the expanded forms equal to each other:
\[
x^3 - 6x^2 + 9x - 1 = x^3 + 6x^2 + 12x + 8
\]

5. Move all terms to one side to set the equation to zero:
\[
x^3 - 6x^2 + 9x - 1 - x^3 - 6x^2 - 12x - 8 = 0
\]
\[
-12x^2 - 3x - 9 = 0
\]

6. Simplify the equation:
\[
-12x^2 - 3x - 9 = 0
\]

7. Divide by -3 to simplify:
\[
4x^2 + x + 3 = 0
\]

8. Solve the quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 1\), and \(c = 3\):
\[
x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}
\]
\[
x = \frac{-1 \pm \sqrt{1 - 48}}{8}
\]
\[
x = \frac{-1 \pm \sqrt{-47}}{8}
\]

Since the discriminant (\(-47\)) is negative, the quadratic equation has no real solutions.

### Summary of Solutions:
- For the first equation \((x+4)^2 - (x-5)^2 = 5(x-4)\), the solution is \(x = -\frac{11}{13}\).
- For the second equation \(x(3-x)^2 - 1 = (x+2)^3\), there are no real solutions.