To solve the equations \( 2 |3x - 1| + 1 = 52 \) and \( |3x - 1| + 1 = 5 \), we'll tackle them one at a time.

### Equation 1:
\[ 2 |3x - 1| + 1 = 52 \]

1. Subtract 1 from both sides:
\[ 2 |3x - 1| = 51 \]

2. Divide by 2:
\[ |3x - 1| = \frac{51}{2} = 25.5 \]

3. Remove the absolute value, giving us two cases:
\[
3x - 1 = 25.5 \quad \text{or} \quad 3x - 1 = -25.5
\]

**Case 1:**
\[
3x - 1 = 25.5
\]
Add 1 to both sides:
\[
3x = 26.5
\]
Divide by 3:
\[
x = \frac{26.5}{3} \approx 8.833
\]

**Case 2:**
\[
3x - 1 = -25.5
\]
Add 1 to both sides:
\[
3x = -24.5
\]
Divide by 3:
\[
x = \frac{-24.5}{3} \approx -8.167
\]

Thus, the solutions for the first equation are:
\[
x \approx 8.833 \quad \text{and} \quad x \approx -8.167
\]

### Equation 2:
\[ |3x - 1| + 1 = 5 \]

1. Subtract 1 from both sides:
\[ |3x - 1| = 4 \]

2. Remove the absolute value, giving us two cases:
\[
3x - 1 = 4 \quad \text{or} \quad 3x - 1 = -4
\]

**Case 1:**
\[
3x - 1 = 4
\]
Add 1 to both sides:
\[
3x = 5
\]
Divide by 3:
\[
x = \frac{5}{3} \approx 1.667
\]

**Case 2:**
\[
3x - 1 = -4
\]
Add 1 to both sides:
\[
3x = -3
\]
Divide by 3:
\[
x = -1
\]

Thus, the solutions for the second equation are:
\[
x = \frac{5}{3} \quad \text{and} \quad x = -1
\]

### Summary of Solutions:
- From the first equation: \( x \approx 8.833 \) and \( x \approx -8.167 \)
- From the second equation: \( x = \frac{5}{3} \) and \( x = -1 \)