(a) To find \(\int_{1}^{\infty} f(x) \, dx\), we first integrate \(f(x) = \frac{2}{(x+2)^2}\).

The integral of \(\frac{2}{(x+2)^2}\) is \(\frac{-2}{x+2}\).

Evaluate the definite integral from 1 to \(\infty\):

\(\left[ \frac{-2}{x+2} \right]_1^{\infty} = 0 - \left( \frac{-2}{3} \right) = \frac{2}{3}\).

(b) Given \(\frac{dy}{dx} = \frac{2}{(x+2)^2}\), integrate to find \(y\):

\(y = \frac{-2}{x+2} + c\).

Using the point \((-1, -1)\), substitute \(x = -1\) and \(y = -1\):

\(-1 = \frac{-2}{-1+2} + c\).

\(-1 = -2 + c\).

\(c = 1\).

Thus, the equation of the curve is \(y = \frac{-2}{x+2} + 1\).