Let's solve the given equations one by one:

d) \((x + \frac{1}{2})^3 : 3 = -\frac{1}{81}\)

To solve this, multiply both sides by 3 to isolate the cube:

\[
(x + \frac{1}{2})^3 = -\frac{1}{27}
\]

Take the cube root of both sides:

\[
x + \frac{1}{2} = -\frac{1}{3}
\]

Subtract \(\frac{1}{2}\) from both sides:

\[
x = -\frac{1}{3} - \frac{1}{2} = -\frac{2}{6} - \frac{3}{6} = -\frac{5}{6}
\]

e) \((2x - \frac{1}{5})^2 + \frac{16}{2} = 1\)

First, simplify \(\frac{16}{2}\) to get 8:

\[
(2x - \frac{1}{5})^2 + 8 = 1
\]

Subtract 8 from both sides:

\[
(2x - \frac{1}{5})^2 = -7
\]

Since a square cannot be negative, there are no real solutions for this equation.

g) The equation appears to be incomplete as given. If you intended to write:

\((8x - 2)^2x + 1 = 5^{2x} + 1\), the equation becomes quite complex to solve analytically without specific context or constraints. If further clarification or constraints are provided, I can assist accordingly.

h) \((2x - 1)^{2026} = (2x \cdot 1)^{2025}\)

This can be simplified to:

\((2x - 1)^{2026} = (2x)^{2025}\)

If \((2x - 1)^{2026} = (2x)^{2025}\), we need to balance the exponents and terms for this to be true:

- This can either mean that \(2x - 1 = 2x\), which implies \(x\) is undefined since subtracting \(2x\) from both sides results in \(-1 = 0\), which is impossible.

Given that straightforward solutions for \(x\) result in contradictions or require complex reasoning, reviewing the original question context or additional constraints could provide further insights.