To solve the equations presented in the image, let’s tackle each part step-by-step.

### a)
Given:
1. \(\frac{x}{5} = \frac{y}{-3}\)
2. \(x^2 + y^2 = 34\)

From the first equation:
\[
x = \frac{5y}{-3} \implies x = -\frac{5y}{3}
\]

Substituting \(x\) into the second equation:
\[
\left(-\frac{5y}{3}\right)^2 + y^2 = 34
\]
\[
\frac{25y^2}{9} + y^2 = 34
\]
\[
\frac{25y^2 + 9y^2}{9} = 34
\]
\[
\frac{34y^2}{9} = 34
\]
\[
34y^2 = 306 \implies y^2 = 9 \implies y = 3 \text{ or } y = -3
\]

Substituting back to find \(x\):
1. If \(y = 3\):
\[
x = -\frac{5(3)}{3} = -5
\]
2. If \(y = -3\):
\[
x = -\frac{5(-3)}{3} = 5
\]

So, the solutions for \(a\) are: \((-5, 3)\) and \((5, -3)\).

### b)
Given:
1. \(\frac{x}{4} = \frac{y}{7}\)
2. \(x^2 - y^2 = -33\)

From the first equation:
\[
x = \frac{4y}{7}
\]

Substituting into the second equation:
\[
\left(\frac{4y}{7}\right)^2 - y^2 = -33
\]
\[
\frac{16y^2}{49} - y^2 = -33
\]
\[
\frac{16y^2 - 49y^2}{49} = -33
\]
\[
\frac{-33y^2}{49} = -33 \implies 33y^2 = 1617 \implies y^2 = 49 \implies y = 7 \text{ or } y = -7
\]

Substituting back to find \(x\):
1. If \(y = 7\):
\[
x = \frac{4(7)}{7} = 4
\]
2. If \(y = -7\):
\[
x = \frac{4(-7)}{7} = -4
\]

So, the solutions for \(b\) are: \((4, 7)\) and \((-4, -7)\).

### c)
Given:
1. \(4x = 3y\)
2. \(x^2 + y^2 = 100\)

From the first equation:
\[
y = \frac{4x}{3}
\]

Substituting into the second equation:
\[
x^2 + \left(\frac{4x}{3}\right)^2 = 100
\]
\[
x^2 + \frac{16x^2}{9} = 100
\]
\[
\frac{9x^2 + 16x^2}{9} = 100
\]
\[
\frac{25x^2}{9} = 100 \implies 25x^2 = 900 \implies x^2 = 36 \implies x = 6 \text{ or } x = -6
\]

Substituting back to find \(y\):
1. If \(x = 6\):
\[
y = \frac{4(6)}{3} = 8
\]
2. If \(x = -6\):
\[
y = \frac{4(-6)}{3} = -8
\]

So, the solutions for \(c\) are: \((6, 8)\) and \((-6, -8)\).

### d)
Given:
1. \(2x = 5y\)
2. \(x^3 + y^3 = 133\)

From the first equation:
\[
y = \frac{2x}{5}
\]

Substituting into the second equation:
\[
x^3 + \left(\frac{2x}{5}\right)^3 = 133
\]
\[
x^3 + \frac{8x^3}{125} = 133
\]
\[
\frac{125x^3 + 8x^3}{125} = 133
\]
\[
\frac{133x^3}{125} = 133 \implies 133x^3 = 16625 \implies x^3 = 125 \implies x = 5
\]

Substituting back to find \(y\):
\[
y = \frac{2(5)}{5} = 2
\]

So, the solution for \(d\) is: \((5, 2)\).

### Summary:
1. a) \((-5, 3)\) and \((5, -3)\)
2. b) \((4, 7)\) and \((-4, -7)\)
3. c) \((6, 8)\) and \((-6, -8)\)
4. d) \((5, 2)\)