(a) The polar equation is \(r = a(\cos \theta + \sin \theta)\). Using \(x = r\cos \theta\) and \(y = r\sin \theta\), we have:

\(x = a\cos \theta (\cos \theta + \sin \theta) = a(\cos^2 \theta + \cos \theta \sin \theta)\)

\(y = a\sin \theta (\cos \theta + \sin \theta) = a(\sin \theta \cos \theta + \sin^2 \theta)\)

Adding these, \(x + y = a(\cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta) = a\).

Thus, \(x + y = a\). The Cartesian equation is \(x^2 + y^2 = ax + ay\), which simplifies to \((x - \frac{a}{2})^2 + (y - \frac{a}{2})^2 = (\frac{a}{\sqrt{2}})^2\).

(b) The greatest distance from the pole is when \(\theta = \frac{\pi}{4}\), giving \(r = a\sqrt{2}\).

(c) To verify \(1.25 < \phi < 1.26\), solve \(\cos \phi + \sin \phi - \phi = 0\). Check values: \(\cos 1.25 + \sin 1.25 - 1.25 = 0.01\) and \(\cos 1.26 + \sin 1.26 - 1.26 = -0.002\), showing a sign change.

(d) The area of the smaller region is:

\(\frac{1}{2} \int_0^\phi \theta^2 \, d\theta + \frac{a^2}{2} \int_\phi^{\frac{3}{4}\pi} (\cos \theta + \sin \theta)^2 \, d\theta\)

Calculate each integral and simplify to get:

\(\frac{1}{2}a^2 \left( \frac{3}{4}\pi + \frac{1}{3}\phi^3 - \phi + \frac{1}{2}\cos 2\phi \right)\)

The area of the larger region is:

\(\frac{a^2}{2} \left( \frac{\pi}{3} + \frac{1}{3}\phi - \frac{1}{2}\cos 2\phi \right)\)