$\lim _{x \rightarrow 0} \frac{\left(7^x-1\right)^4}{\tan \left(\frac{x}{\mathrm{k}}\right) \cdot \log \left(1+\frac{x^2}{3}\right) \cdot \sin 4 x}=3(\log 7)^3$

Evaluate the limit $L = \lim_{x \to 0} \frac{(7^x - 1)^4}{\tan \left(\frac{x}{k}\right) \cdot \log \left(1 + \frac{x^2}{3}\right) \cdot \sin 4x}$ given that it equals $3(\log 7)^3$.

Apply standard limits $\lim_{x \to 0} \frac{7^x - 1}{x} = \log 7$, $\lim_{x \to 0} \frac{\tan x}{x} = 1$, $\lim_{x \to 0} \frac{\log(1 + x)}{x} = 1$, and $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

Express each component in terms of these limits:
Numerator: $(7^x - 1)^4 = \left(\frac{7^x - 1}{x}\right)^4 x^4 \to (\log 7)^4 x^4$.
Denominator terms:
$\tan \left(\frac{x}{k}\right) = \frac{\tan(x/k)}{x/k} \cdot \frac{x}{k} \to \frac{x}{k}$,
$\log(1 + x^2/3) = \frac{\log(1 + x^2/3)}{x^2/3} \cdot \frac{x^2}{3} \to \frac{x^2}{3}$,
$\sin 4x = \frac{\sin 4x}{4x} \cdot 4x \to 4x$.

Substitute into $L$:
$L = \lim_{x \to 0} \frac{(\log 7)^4 x^4}{(x/k)(x^2/3)(4x)} = \frac{(\log 7)^4 x^4}{(4x^4)/(3k)} = \frac{3k (\log 7)^4}{4}$.

Set equal to given value:
$\frac{3k (\log 7)^4}{4} = 3(\log 7)^3$.
Divide both sides by $3(\log 7)^3$: $\frac{k \log 7}{4} = 1$ so $k = \frac{4}{\log 7} = 4(\log 7)^{-1}$.

This matches option A.