Sure, let's solve these equations one by one.

### Equation b):
\[
\frac{x+6}{5} - \frac{x-2}{3} = 2
\]

To solve this, find a common denominator (which is 15):

\[
\frac{3(x+6) - 5(x-2)}{15} = 2
\]

Simplify the numerator:

\[
3(x+6) - 5(x-2) = 3x + 18 - 5x + 10 = -2x + 28
\]

Now the equation becomes:

\[
\frac{-2x + 28}{15} = 2
\]

Multiply both sides by 15:

\[
-2x + 28 = 30
\]

Solve for \(x\):

\[
-2x = 30 - 28
\]

\[
-2x = 2
\]

\[
x = -1
\]

### Equation b):
\[
\frac{1}{x+2} + \frac{5}{x-2} = \frac{2x-12}{x^2-4}
\]

Notice that \(x^2 - 4 = (x+2)(x-2)\). Therefore, the equation can be written as:

\[
\frac{1}{x+2} + \frac{5}{x-2} = \frac{2x-12}{(x+2)(x-2)}
\]

Combine the left side with a common denominator:

\[
\frac{(x-2) + 5(x+2)}{(x+2)(x-2)} = \frac{2x-12}{(x+2)(x-2)}
\]

Simplify the numerator on the left:

\[
x - 2 + 5x + 10 = 6x + 8
\]

Now the equation becomes:

\[
\frac{6x + 8}{(x+2)(x-2)} = \frac{2x-12}{(x+2)(x-2)}
\]

Since the denominators are the same, set the numerators equal:

\[
6x + 8 = 2x - 12
\]

Solve for \(x\):

\[
6x - 2x = -12 - 8
\]

\[
4x = -20
\]

\[
x = -5
\]

### Equation f):
\[
\frac{6x-2}{15} = 1 - \frac{3-4x}{9}
\]

Subtract the fraction on the right from 1:

\[
1 - \frac{3-4x}{9} = \frac{9 - (3-4x)}{9} = \frac{9 - 3 + 4x}{9} = \frac{4x + 6}{9}
\]

Now the equation is:

\[
\frac{6x-2}{15} = \frac{4x+6}{9}
\]

Cross-multiply to solve for \(x\):

\[
9(6x-2) = 15(4x+6)
\]

Expand both sides:

\[
54x - 18 = 60x + 90
\]

Rearrange and solve for \(x\):

\[
54x - 60x = 90 + 18
\]

\[
-6x = 108
\]

\[
x = -18
\]

So the solutions are:
- \(x = -1\) for the first equation,
- \(x = -5\) for the second equation,
- \(x = -18\) for the third equation.