(a) To find k when \angle BAC = \frac{1}{6}\pi radians:

The area of triangle \Delta ADE is given by:

\(\frac{1}{2} (ka)^2 \sin \frac{\pi}{6} = \frac{1}{4} k^2 a^2\)

The area of sector ABC is:

\(\frac{1}{2} a^2 \frac{\pi}{6}\)

Since DE divides the sector into two equal areas:

\(2 \times \frac{1}{4} k^2 a^2 = \frac{1}{2} a^2 \frac{\pi}{6}\)

Simplifying gives:

\(k^2 = \frac{\pi}{6}\)

\(k = \sqrt{\frac{\pi}{6}} \approx 0.7236\)

(b) For the general case with \angle BAC = \theta:

The area of triangle \Delta ADE is:

\(\frac{1}{2} (ka)^2 \sin \theta\)

Equating twice the area of \Delta ADE to the area of sector ABC:

\(2 \times \frac{1}{2} (ka)^2 \sin \theta = \frac{1}{2} a^2 \theta\)

\(k^2 = \frac{\theta}{2 \sin \theta}\)

Given \(\frac{\theta}{\sin \theta} > 1\), it follows:

\(k^2 > \frac{1}{2}\)

Thus, \(\frac{1}{\sqrt{2}} < k < 1\)