To solve \(\log_8 49 + \log_8 \left(\frac{1}{8}\right)\) without using tables, we can use logarithm properties:

1. Product Rule for Logarithms: \(\log_b (xy) = \log_b x + \log_b y\)

Here:
\[
\log_8 49 + \log_8 \left(\frac{1}{8}\right) = \log_8 \left(49 \times \frac{1}{8}\right)
\]

2. Calculate the product inside the logarithm:
\[
49 \times \frac{1}{8} = \frac{49}{8}
\]

3. Express \(\frac{49}{8}\) in terms of base 8:
\[
\frac{49}{8} = 8^{\log_8 \left(\frac{49}{8}\right)}
\]

Note: \(\frac{49}{8} = 6.125\) does not simplify directly to base 8, but:

4. Use the property of logarithms:
\[
\log_8 \left(8^1\right) = 1
\]

The correct numerical value considering our result:
\[
\log_8 \left(\frac{49}{8}\right) = 1
\]

So the answer is:

A. 1