To solve the equations, we can approach them by considering the cases determined by the critical points where the expressions inside the absolute values change sign.

### Equation (b): \( |x-2| + |x-5| = 3 \)

**Critical Points:**
- \( x = 2 \)
- \( x = 5 \)

**Case 1:** \( x < 2 \)
- Here, \( |x-2| = 2 - x \) and \( |x-5| = 5 - x \).
- Equation becomes:
\[
(2 - x) + (5 - x) = 3 \implies 7 - 2x = 3 \implies 2x = 4 \implies x = 2
\]
- This value does not satisfy \( x < 2 \).

**Case 2:** \( 2 \leq x < 5 \)
- Here, \( |x-2| = x - 2 \) and \( |x-5| = 5 - x \).
- Equation becomes:
\[
(x - 2) + (5 - x) = 3 \implies 3 = 3
\]
- This is true for all \( x \) in the interval \( [2, 5) \).

**Case 3:** \( x \geq 5 \)
- Here, \( |x-2| = x - 2 \) and \( |x-5| = x - 5 \).
- Equation becomes:
\[
(x - 2) + (x - 5) = 3 \implies 2x - 7 = 3 \implies 2x = 10 \implies x = 5
\]
- This value satisfies \( x \geq 5 \).

**Solution for (b):** \( x \in [2, 5] \) (including endpoints).

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### Equation (c): \( |3x-5| + |3x+1| = 6 \)

**Critical Points:**
- \( 3x - 5 = 0 \) implies \( x = \frac{5}{3} \)
- \( 3x + 1 = 0 \) implies \( x = -\frac{1}{3} \)

**Case 1:** \( x < -\frac{1}{3} \)
- Here, \( |3x-5| = 5 - 3x \) and \( |3x+1| = -3x - 1 \).
- Equation becomes:
\[
(5 - 3x) + (-3x - 1) = 6 \implies 4 - 6x = 6 \implies -6x = 2 \implies x = -\frac{1}{3}
\]
- This does not satisfy \( x < -\frac{1}{3} \).

**Case 2:** \( -\frac{1}{3} \leq x < \frac{5}{3} \)
- Here, \( |3x-5| = 5 - 3x \) and \( |3x+1| = 3x + 1 \).
- Equation becomes:
\[
(5 - 3x) + (3x + 1) = 6 \implies 6 = 6
\]
- This is true for all \( x \) in the interval \( \left[-\frac{1}{3}, \frac{5}{3}\right) \).

**Case 3:** \( x \geq \frac{5}{3} \)
- Here, \( |3x-5| = 3x - 5 \) and \( |3x+1| = 3x + 1 \).
- Equation becomes:
\[
(3x - 5) + (3x + 1) = 6 \implies 6x - 4 = 6 \implies 6x = 10 \implies x = \frac{5}{3}
\]
- This satisfies \( x \geq \frac{5}{3} \).

**Solution for (c):** \( x \in \left[-\frac{1}{3}, \frac{5}{3}\right] \) (including endpoints).

### Final Solutions:
- (b) \( x \in [2, 5] \)
- (c) \( x \in \left[-\frac{1}{3}, \frac{5}{3}\right] \)