To find \( x \), let's solve each part of the problem:

### a) \(\frac{x-2}{3} - \frac{x-2}{4} + \frac{x-2}{5} = 0\)

First, notice that the common term is \( x-2 \). This allows us to factor it out:

\[
(x-2) \left(\frac{1}{3} - \frac{1}{4} + \frac{1}{5}\right) = 0
\]

Since a product is zero if and only if one of the factors is zero, we have:

1. \( x-2 = 0 \Rightarrow x = 2 \)
2. The other possibility is to solve \(\frac{1}{3} - \frac{1}{4} + \frac{1}{5} = 0\).

To solve \(\frac{1}{3} - \frac{1}{4} + \frac{1}{5}\), find a common denominator:

\[
\frac{1}{3} - \frac{1}{4} + \frac{1}{5} = \frac{20}{60} - \frac{15}{60} + \frac{12}{60} = \frac{17}{60}
\]

Since \(\frac{17}{60} \neq 0\), the only solution is \( x = 2 \).

### b) \(\frac{x+1}{15} + \frac{x+2}{14} - \frac{x+3}{13} + \frac{x+4}{12}\)

We need to solve this equation:

To simplify further, we would equate the expression to zero and find a common denominator for \(15, 14, 13, 12\). However, without more context, we cannot directly simplify this unless there is more specific information or context.

Given the setup, \(x\) can be solved numerically or symbolically depending on further transformation or information on the equation. If the sum equals zero:

1. Combine terms over a common denominator.
2. Solve for \(x\) from the equation with equal denominators.

Without explicit bounds for parameters or additional equation setup, indicating the complete solution for \(x\) specifically depends on further constraints.

For precise calculation, solving either by substitution and algebra or a numeric method (calculator/computer-based) might be appropriate approaches.