(i) To find the stationary point, we need to find \(\frac{dy}{dx}\) and set it to zero. Using the quotient rule:

\(\frac{dy}{dx} = \frac{-3 \sin x (2 + \sin x) - 3 \cos x \cos x}{(2 + \sin x)^2}\)

Equate the numerator to zero:

\(-3 \sin x (2 + \sin x) - 3 \cos^2 x = 0\)

\(-6 \sin x - 3 \sin^2 x - 3 (1 - \sin^2 x) = 0\)

\(-6 \sin x - 3 + 3 \sin^2 x = 0\)

\(3 \sin^2 x + 6 \sin x + 3 = 0\)

\(\sin x = -\frac{1}{2}\)

\(x = -\frac{\pi}{6}\)

Substitute back to find \(y\):

\(y = \frac{3 \cos(-\frac{\pi}{6})}{2 + \sin(-\frac{\pi}{6})} = \sqrt{3}\)

(ii) To find \(a\), we evaluate the integral:

\(\int \frac{3 \cos x}{2 + \sin x} \, dx = 3 \ln |2 + \sin x| + C\)

Substitute limits and equate to 1:

\(3 \ln(2 + \sin a) - 3 \ln 2 = 1\)

\(3 \ln \left( \frac{2 + \sin a}{2} \right) = 1\)

\(\ln \left( \frac{2 + \sin a}{2} \right) = \frac{1}{3}\)

\(\frac{2 + \sin a}{2} = e^{\frac{1}{3}}\)

\(2 + \sin a = 2e^{\frac{1}{3}}\)

\(\sin a = 2e^{\frac{1}{3}} - 2\)

\(a = \sin^{-1}(2e^{\frac{1}{3}} - 2) \approx 0.913\)