To find the sum of the intercepts of the line represented by the equation \(\frac{x}{a^2} + \frac{y}{b^2} = \frac{2}{a^2 + b^2}\) on the coordinate axes, we can follow these steps:

1. Identify the form of the linear equation: The given equation is not in the standard linear form. We need to convert it into the intercept form of a line which is: \(\frac{x}{A} + \frac{y}{B} = 1\), where \(A\) and \(B\) are the x-intercept and y-intercept respectively.
2. Rewrite the equation:

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

1. To match the intercept form, multiply both sides of the equation by \(a^2 + b^2\):

\((a^2 + b^2) \left(\frac{x}{a^2} + \frac{y}{b^2}\right) = 2\)

1. Simplify the equation:

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

\(\frac{a^2 + b^2}{a^2} \cdot x + \frac{a^2 + b^2}{b^2} \cdot y = 2\)

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

1. Rearrange it to:

\(\left(\frac{x}{\frac{a^2}{b^2}}\right) + \left(\frac{y}{\frac{b^2}{a^2}}\right) = 2\)

1. Determine the intercepts: The intercepts \()-intercept \(A = a^2/b^2\) and \(y\)-intercept \(B = b^2/a^2\) are obtained from the equatio\)
2. Calculate the sum of intercepts:

The sum \(S\) of intercepts \(A\) and \(B\) is given by:

\(S = A + B = \frac{a^2}{b^2} + \frac{b^2}{a^2}\)

Express as a single fraction: \[ S = \frac{a^4 + b^4}{a^2 b^2} \]

Recognize this can be rewritten using identity: \[ a^4 + b^4 = (a^2 + b^2)^2 - 2(a^2)(b^2) \] thus:

\[ S = \frac{(a^2 + b^2)^2 - 2a^2b^2}{a^2b^2} = \frac{(a^2 + b^2)^2}{a^2b^2} - 2 \]

Using the fact that \(\frac{(a^2 + b^2)^2}{a^2b^2} = 2\), the final result is:

\(S = 2\)

Therefore, the sum of the intercepts of the line on the coordinate axes is 2.