Let's solve each of the inequalities one by one.

**a.** Solve \((x-1)(x+2) > (x-1)^2 + 3\):

1. Expand both sides:
\[
(x-1)(x+2) = x^2 + 2x - x - 2 = x^2 + x - 2
\]
\[
(x-1)^2 + 3 = x^2 - 2x + 1 + 3 = x^2 - 2x + 4
\]

2. Set up the inequality:
\[
x^2 + x - 2 > x^2 - 2x + 4
\]

3. Subtract \(x^2\) from both sides:
\[
x - 2 > -2x + 4
\]

4. Bring all terms involving \(x\) on one side:
\[
x + 2x > 4 + 2
\]
\[
3x > 6
\]
\[
x > 2
\]

Thus, the solution for part **a** is:
\[
x > 2
\]

---

**b.** Solve \(x(2x-1) - 8 < 5 - 2x(1-x)\):

1. Simplifying the right-hand side:
\[
5 - 2x(1-x) = 5 - 2x + 2x^2
\]

2. Set up the inequality:
\[
x(2x - 1) - 8 < 5 - 2x + 2x^2
\]
Which expands to:
\[
2x^2 - x - 8 < 5 - 2x + 2x^2
\]

3. Subtract \(2x^2\) from both sides:
\[
-x - 8 < 5 - 2x
\]

4. Rearranging yields:
\[
-x + 2x < 5 + 8
\]
\[
x < 13
\]

Thus, the solution for part **b** is:
\[
x < 13
\]

---

**c.** Solve \((2x+1)^2 + (1-x)3x \leq (x+2)^2\):

1. Expanding both sides gives us:
\[
(2x + 1)^2 = 4x^2 + 4x + 1
\]
\[
(1 - x)3x = 3x - 3x^2
\]
\[
(x + 2)^2 = x^2 + 4x + 4
\]

2. Now set up the inequality:
\[
4x^2 + 4x + 1 + 3x - 3x^2 \leq x^2 + 4x + 4
\]

3. Combining terms:
\[
(4x^2 - 3x^2 - x^2) + (4x + 3x - 4x) + 1 \leq 4
\]
Simplifying gives:
\[
0 \leq 3
\]
This is true for all \(x\).

Thus, the solution for part **c** is:
\[
\text{All } x.
\]

---

**d.** Solve \((x-4)(x+4) \geq (x+3)^2 + 5\):

1. Expanding both sides:
\[
(x-4)(x+4) = x^2 - 16
\]
\[
(x+3)^2 + 5 = x^2 + 6x + 9 + 5 = x^2 + 6x + 14
\]

2. Set up the inequality:
\[
x^2 - 16 \geq x^2 + 6x + 14
\]

3. Subtract \(x^2\) from both sides:
\[
-16 \geq 6x + 14
\]

4. Rearranging gives:
\[
-16 - 14 \geq 6x
\]
\[
-30 \geq 6x
\]
\[
x \leq -5
\]

Thus, the solution for part **d** is:
\[
x \leq -5
\]

---

In summary:

- **a:** \(x > 2\)
- **b:** \(x < 13\)
- **c:** All \(x\)
- **d:** \(x \leq -5\)