Correct Answer: A (2)

Note on the problem statement: The question appears to contain two typos.

1. The exponential equation $8^{z-1} = (\frac{1}{4})^x$ likely intended the variable to be $z$ on both sides, i.e., $8^{z-1} = (\frac{1}{4})^z$.
2. For the value of $z$ obtained from this equation, the logarithmic base $(z-1)$ is negative, which is undefined for real logarithms. The base was likely intended to be $(1-z)$.

Step 1: Solve for $z$ from the Corrected Exponential Equation

We start with the corrected equation and express both sides with a common base of 2.

$$ 8^{z-1} = \left(\frac{1}{4}\right)^z $$

$$ (2^3)^{z-1} = (2^{-2})^z $$

$$ 2^{3(z-1)} = 2^{-2z} $$

Since the bases are equal, we can equate the exponents:

$3(z-1) = -2z$
$3z - 3 = -2z$
$5z = 3$
$z = \frac{3}{5}$

Step 2: Simplify the Corrected Logarithmic Expression

The original expression has an undefined term $\log_{z-1}$ because $z-1 = 3/5 - 1 = -2/5$. We assume the expression was meant to be:

$$ \frac{1}{\log_{z+1} 4 - \log_{z+1} 5} + \frac{1}{\log_{1-z} 4 - \log_{1-z} 5} $$

First, we simplify the denominators using the logarithm property $\log_b M - \log_b N = \log_b(\frac{M}{N})$:

$$ \frac{1}{\log_{z+1}(4/5)} + \frac{1}{\log_{1-z}(4/5)} $$

Next, we use the change of base property $\frac{1}{\log_b a} = \log_a b$:

$$ \log_{4/5}(z+1) + \log_{4/5}(1-z) $$

Finally, we combine the terms using the property $\log_b M + \log_b N = \log_b(MN)$:

$$ \log_{4/5}((z+1)(1-z)) = \log_{4/5}(1-z^2) $$

Step 3: Substitute the Value of $z$

Now we substitute our calculated value $z = \frac{3}{5}$ into the simplified expression:

$$ \log_{4/5}\left(1 - \left(\frac{3}{5}\right)^2\right) $$

$$ = \log_{4/5}\left(1 - \frac{9}{25}\right) $$

$$ = \log_{4/5}\left(\frac{16}{25}\right) $$

We can rewrite the argument as a power of the base:

$$ = \log_{4/5}\left(\left(\frac{4}{5}\right)^2\right) $$

Using the property $\log_b(b^p) = p$, we get the final answer:

$$ = 2 $$