Sure, here are the solutions for the equations:

a) \(4x^2 - 36 = 0\)

\[4x^2 = 36\]

\[x^2 = 9\]

\[x = \pm 3\]

b) \(25x^2 - (3x + 1)^2 = 0\)

\[25x^2 = (3x + 1)^2\]

Solve by expanding and equating:

\(25x^2 = 9x^2 + 6x + 1\)

\[16x^2 - 6x - 1 = 0\]

Using the quadratic formula:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

With \(a = 16\), \(b = -6\), \(c = -1\):

\[x = \frac{6 \pm \sqrt{36 + 64}}{32}\]

\[x = \frac{6 \pm 10}{32}\]

\[x = \frac{16}{32} = \frac{1}{2} \,\, \text{or} \,\, x = \frac{-4}{32} = -\frac{1}{8}\]

c) \(x^2 + 12x + 36 = 0\)

\((x + 6)^2 = 0\)

\[x = -6\]

d) \(18x^2 + 12x + 2 = 0\)

Using the quadratic formula:

\[x = \frac{-12 \pm \sqrt{12^2 - 4 \times 18 \times 2}}{2 \times 18}\]

\[x = \frac{-12 \pm \sqrt{144 - 144}}{36}\]

\[x = \frac{-12}{36} = -\frac{1}{3}\]

e) \(4x^2 - 4x - 1 = 0\)

Using the quadratic formula:

\[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 4 \times (-1)}}{2 \times 4}\]

\[x = \frac{4 \pm \sqrt{16 + 16}}{8}\]

\[x = \frac{4 \pm \sqrt{32}}{8}\]

\[x = \frac{4 \pm 4\sqrt{2}}{8}\]

\[x = \frac{1 \pm \sqrt{2}}{2}\]

f) \(8x^4 - 18x^2 = -1\)

Set it to zero and solve:

\[8x^4 - 18x^2 + 1 = 0\]

Set \(y = x^2\), solving:

\[8y^2 - 18y + 1 = 0\]

Use the quadratic formula for \(y\) and back-substitute for \(x\).

g) \(x^3 + 6x^2 + 12x + 8 = 0\)

Factorization or synthetic division might be useful to simplify.

h) \(8x^3 + 12x^2 + 6x + 2 = 0\)

Factorization or synthetic division might be useful to simplify.

i) \(27x^3 - 27x^2 + 9x = 1\)

First move all terms to one side:

\[27x^3 - 27x^2 + 9x - 1 = 0\]

Factorization or synthetic division might be useful to attempt simplification.

j) \(125x^3 - 150x^2 = 9 - 60x\)

\[125x^3 - 150x^2 + 60x - 9 = 0\]

Try factoring or using numeric/root-solving techniques to get possible real solutions.