If \(\alpha\) and \(\beta\) are acute angles such that \(\alpha - \beta\) and \(\alpha + \beta\) satisfy the equation \(\tan^2\theta - 4\tan\theta + 1 = 0\), then \(\alpha\) and \(\beta\) are respectively,

The roots of the equation \(\tan^2\theta - 4\tan\theta + 1 = 0\) are \(\tan(\alpha - \beta)\) and \(\tan(\alpha + \beta)\).

Sum of roots: \(\tan(\alpha - \beta) + \tan(\alpha + \beta) = 4\)

Product of roots: \(\tan(\alpha - \beta) \tan(\alpha + \beta) = 1\)

Using the tangent addition formula:

\(\tan(2\alpha) = \tan((\alpha - \beta) + (\alpha + \beta)) = \dfrac{\tan(\alpha - \beta) + \tan(\alpha + \beta)}{1 - \tan(\alpha - \beta) \tan(\alpha + \beta)}\)

Substituting the sum and product of roots:

\(\tan(2\alpha) = \dfrac{4}{1 - 1} \rightarrow \infty\)

Since \(\alpha\) is an acute angle, \(2\alpha = 90^\circ \Rightarrow \alpha = 45^\circ\).

Now, substituting \(\alpha = 45^\circ\) into the sum of roots equation:

\(\tan(45^\circ - \beta) + \tan(45^\circ + \beta) = 4\)

\(\dfrac{1 - \tan\beta}{1 + \tan\beta} + \dfrac{1 + \tan\beta}{1 - \tan\beta} = 4\)

\(\dfrac{(1 - \tan\beta)^2 + (1 + \tan\beta)^2}{1 - \tan^2\beta} = 4\)

\(\dfrac{2(1 + \tan^2\beta)}{1 - \tan^2\beta} = 4\)

\(1 + \tan^2\beta = 2 - 2\tan^2\beta\)

\(3\tan^2\beta = 1 \Rightarrow \tan\beta = \dfrac{1}{\sqrt{3}}\) (since \(\beta\) is acute)

Thus, \(\beta = 30^\circ\).

Answer: \(45^\circ, 30^\circ\)