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Solution

Let \( \tan^{-1}x=\theta \).
Then \( \cot^{-1}x=\frac{\pi}{2}-\theta \).
Since \( x\in R \), \( -\frac{\pi}{2}<\theta<\frac{\pi}{2} \).

Given,
\( (\tan^{-1}x)^3+(\cot^{-1}x)^3=k\pi^3 \)
\( \theta^3+\left(\frac{\pi}{2}-\theta\right)^3=k\pi^3 \)

Let \( t=\frac{\theta}{\pi} \).
So, $\dfrac{-1}{2}<t<\dfrac{1}{2}$
Now,
\( k=t^3+\left(\frac{1}{2}-t\right)^3 \)
\( k=t^3+\frac{1}{8}-\frac{3t}{4}+\frac{3t^2}{2}-t^3 \)
\( k=\frac{3t^2}{2}-\frac{3t}{4}+\frac{1}{8} \)

This quadratic is minimum at \( t=\frac{1}{4} \).
\( k_{\min}=\frac{1}{32} \)

As \( t\to-\frac{1}{2} \), \( k\to\frac{7}{8} \), but \( t=-\frac{1}{2} \) is not included.

Therefore,
\( k\in\left[\frac{1}{32},\frac{7}{8}\right) \)

Answer: \( \boxed{(a)\ \left[\frac{1}{32},\frac{7}{8}\right)} \)