To solve the systems of equations provided, let’s break them down one by one.

### System (d):
1. \(\frac{x}{-2} = \frac{y}{5}\)
2. \(x + y = 12\)

From the first equation, we can cross-multiply:

\[
5x = -2y
\]

Rearranging gives:

\[
y = -\frac{5}{2}x
\]

Now, substitute \(y\) into the second equation:

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

Combining like terms results in:

\[
-\frac{3}{2}x = 12
\]

Solving for \(x\):

\[
x = 12 \cdot -\frac{2}{3} = -8
\]

Now substitute \(x\) back into the equation for \(y\):

\[
y = -\frac{5}{2}(-8) = 20
\]

Thus, the solution for system (d) is:
\[
(x, y) = (-8, 20)
\]

### System (b):
1. \(3x = 4y\)
2. \(2x + 3y = 34\)

From the first equation, we isolate \(y\):

\[
y = \frac{3}{4}x
\]

Substituting \(y\) into the second equation:

\[
2x + 3\left(\frac{3}{4}x\right) = 34
\]

Simplifying gives:

\[
2x + \frac{9}{4}x = 34
\]

To eliminate the fraction, multiply the entire equation by 4:

\[
8x + 9x = 136
\]

Combining the \(x\) terms results in:

\[
17x = 136
\]

Solving for \(x\):

\[
x = \frac{136}{17} = 8
\]

Now substitute back to find \(y\):

\[
y = \frac{3}{4}(8) = 6
\]

Thus, the solution for system (b) is:
\[
(x, y) = (8, 6)
\]

### Summary of Solutions:
- For system (d): \((-8, 20)\)
- For system (b): \((8, 6)\)