(a) Let \(u = \cos x\), then \(\frac{du}{dx} = -\sin x\) or \(dx = -\frac{du}{\sin x}\).

Using the double angle formula, \(\sin 2x = 2 \sin x \cos x = 2 \sin x \cdot u\).

Substitute into the integral:

\(\int \sin 2x \, e^{2 \cos x} \, dx = \int 2 \sin x \cdot u \, e^{2u} \cdot -\frac{du}{\sin x} = \int -2u e^{2u} \, du\).

Change the limits: when \(x = 0, u = \cos 0 = 1\) and when \(x = \pi, u = \cos \pi = -1\).

Thus, \(\int_{0}^{\pi} \sin 2x \, e^{2 \cos x} \, dx = \int_{1}^{-1} -2u e^{2u} \, du = \int_{-1}^{1} 2u e^{2u} \, du\).

(b) Integrate \(\int_{-1}^{1} 2u e^{2u} \, du\):

Let \(I = \int 2u e^{2u} \, du\).

Using integration by parts, let \(v = u\) and \(dw = 2e^{2u} \, du\), then \(dv = du\) and \(w = e^{2u}\).

\(I = u e^{2u} - \int e^{2u} \, du = u e^{2u} - \frac{1}{2} e^{2u} + C\).

Evaluate from \(-1\) to \(1\):

\(\left[ u e^{2u} - \frac{1}{2} e^{2u} \right]_{-1}^{1} = \left( 1 \cdot e^{2} - \frac{1}{2} e^{2} \right) - \left( -1 \cdot e^{-2} - \frac{1}{2} e^{-2} \right)\).

\(= \frac{1}{2} e + \frac{3}{2} e^{-2}\).