Evaluate the first logarithm term, 64 256.

We can express both the base and the argument as powers of 2.
$64 = 2^6$ and $256 = 2^8$.
Using the logarithm property $\log_{b^m} a^n = \frac{n}{m} \log_b a$:

$$^{64}\log 256 = \log_{2^6} 2^8 = \frac{8}{6} \log_2 2$$

Since $\log_2 2 = 1$:

$$^{64}\log 256 = \frac{8}{6} \times 1 = \frac{4}{3}$$

We can express both the base and the argument as powers of 3.
$27 = 3^3$ and $9 = 3^2$.
Using the logarithm property $\log_{b^m} a^n = \frac{n}{m} \log_b a$:

$$^{27}\log 9 = \log_{3^3} 3^2 = \frac{2}{3} \log_3 3$$

Since $\log_3 3 = 1$:

$$^{27}\log 9 = \frac{2}{3} \times 1 = \frac{2}{3}$$

$$\frac{4}{3} + \frac{2}{3} = \frac{4+2}{3} = \frac{6}{3} = 2$$

The value of the expression is 2.

The calculated value is 2, which matches option C.

The final answer is $\text{C) 2}$.