(i) We know that \(\sin^2 x + \cos^2 x = 1\).

Substitute \(\sin x = a + b\) and \(\cos x = a - b\):

\((a + b)^2 + (a - b)^2 = 1\)

Expanding both terms, we get:

\(a^2 + 2ab + b^2 + a^2 - 2ab + b^2 = 1\)

Combine like terms:

\(2a^2 + 2b^2 = 1\)

Divide by 2:

\(a^2 + b^2 = \frac{1}{2}\)

(ii) Given \(\tan x = 2\), we have:

\(\tan x = \frac{\sin x}{\cos x} = \frac{a + b}{a - b} = 2\)

Cross-multiply to solve for a:

\(a + b = 2(a - b)\)

\(a + b = 2a - 2b\)

Rearrange terms:

\(b + 2b = 2a - a\)

\(3b = a\)

Thus, \(a = 3b\).