Let's solve each inequality step by step.

### Inequality 1:
\[ 2(x + 2)^2 < 2x(x + 2) + 4 \]

1. Expand both sides:
\[
2(x^2 + 4x + 4) < 2x^2 + 4x + 4
\]
\[
2x^2 + 8x + 8 < 2x^2 + 4x + 4
\]

2. Subtract \(2x^2\) from both sides:
\[
8x + 8 < 4x + 4
\]

3. Rearrange:
\[
8x - 4x < 4 - 8
\]
\[
4x < -4
\]

4. Divide by 4:
\[
x < -1
\]

### Solution for Inequality 1:
\( x < -1 \)

---

### Inequality 2:
\[ (x - 1)^2 + x^2 < (x + 1)^2 + (x + 2)^2 \]

1. Expand both sides:
\[
(x^2 - 2x + 1) + x^2 < (x^2 + 2x + 1) + (x^2 + 4x + 4)
\]
\[
2x^2 - 2x + 1 < 2x^2 + 6x + 5
\]

2. Subtract \(2x^2\) from both sides:
\[
-2x + 1 < 6x + 5
\]

3. Rearrange:
\[
-2x - 6x < 5 - 1
\]
\[
-8x < 4
\]

4. Divide by -8 (remember to reverse the inequality):
\[
x > -\frac{1}{2}
\]

### Solution for Inequality 2:
\( x > -\frac{1}{2} \)

---

### Inequality 3:
\[ (x^2 + 1)(x - 6) < (x - 2)^3 \]

1. Expand the right side:
\[
(x^2 + 1)(x - 6) < (x - 2)(x - 2)(x - 2)
\]
\[
(x^2 + 1)(x - 6) < (x^2 - 4x + 4)(x - 2)
\]

Expanding the right side:
\[
(x^2 - 4x + 4)(x - 2) = x^3 - 2x^2 - 4x^2 + 8x + 4x - 8 = x^3 - 6x^2 + 8x - 8
\]

2. Now we have:
\[
(x^3 - 6x^2 + 8x - 8) - (x^3 - 6x + (x + 1)(x - 6)) > 0
\]

This simplifies to solving:
\[
2x - 6 > 0
\]

Rearranging gives:
\[
2x > 6 \implies x > 3
\]

### Solution for Inequality 3:
\( x > 3 \)

---

### Final Solutions:
1) \( x < -1 \)
2) \( x > -\frac{1}{2} \)
3) \( x > 3 \)