Given \( a < b \), we need to prove the following inequalities.

**a)** \(-2a + 4 > -2b + 4\)

We can rearrange the inequality as follows:
\[
-2a + 4 > -2b + 4 \implies -2a > -2b
\]
Dividing both sides by \(-2\) (and remembering to reverse the inequality):
\[
a < b
\]
This is true since we assumed \( a < b \). Thus, the inequality is proven.

---

**b)** \(5 - 3a > 1 - 3b\)

Rearranging gives:
\[
5 - 3a > 1 - 3b \implies 5 - 1 > 3a - 3b \implies 4 > 3a - 3b
\]
Factoring out the \(3\):
\[
4 > 3(a - b)
\]
Since \( a < b \), we know that \( a - b < 0 \). Thus \( 3(a - b) < 0 \) which leads to:
\[
4 > 0
\]
This is true, so the inequality \( 5 - 3a > 1 - 3b \) is proven.

---

**c)** \(4a + 3^2 < 4b + (-3)^2\)

We can simplify this as:
\[
4a + 9 < 4b + 9
\]
Subtracting \(9\) from both sides:
\[
4a < 4b
\]
Dividing both sides by \(4\):
\[
a < b
\]
Which is true based on our assumption. Therefore, the inequality \(4a + 3^2 < 4b + (-3)^2\) is proven.

---

**d)** \(-a + 1 < -b + 1\)

Rearranging gives:
\[
-a + 1 < -b + 1 \implies -a < -b
\]
Dividing both sides by \(-1\) (again reversing the inequality):
\[
a > b
\]
This seems inconsistent with our assumption \( a < b \). Therefore, we will not prove this inequality \( -a + 1 < -b + 1 \) under the initial condition \( a < b\).

### Summary
The inequalities a, b, and c are proven to be true under the condition \( a < b \). However, part d is actually false and does not hold under the same condition.