(a) To solve the differential equation \(\frac{dy}{dx} = \frac{1 + 4y^2}{e^x}\), we first separate variables:

\(e^x \, dy = (1 + 4y^2) \, dx\).

Integrate both sides:

\(\int \frac{1}{1 + 4y^2} \, dy = \int e^{-x} \, dx\).

The left side integrates to \(\frac{1}{2} \arctan(2y)\) and the right side to \(-e^{-x} + C\).

Thus, \(\frac{1}{2} \arctan(2y) = -e^{-x} + C\).

Using the initial condition \(y = 0\) when \(x = 1\), we find:

\(\frac{1}{2} \arctan(0) = -e^{-1} + C\).

Since \(\arctan(0) = 0\), we have \(0 = -e^{-1} + C\), so \(C = e^{-1}\).

Substitute back to get:

\(\frac{1}{2} \arctan(2y) = -e^{-x} + e^{-1}\).

Solving for \(y\), we get:

\(y = \frac{1}{2} \tan \left( 2e^{-1} - 2e^{-x} \right)\).

(b) As \(x \to \infty\), \(e^{-x} \to 0\), so:

\(y \to \frac{1}{2} \tan \left( 2e^{-1} \right)\).