If $$y = y(x)$$ is the solution of the differential equation $$\frac{dy}{dx} + 2y = \sin(2x), y(0) = \frac{3}{4}$$, then $$y\left(\frac{\pi}{8}\right)$$ is equal to:

$$\frac{dy}{dx} + 2y = \sin 2x$$, $$y(0) = \frac{3}{4}$$.

Integrating factor: $$e^{\int 2dx} = e^{2x}$$.

$$ ye^{2x} = \int e^{2x}\sin 2x \, dx $$

Using the formula $$\int e^{ax}\sin bx \, dx = \frac{e^{ax}(a\sin bx - b\cos bx)}{a^2 + b^2}$$:

With $$a = 2, b = 2$$:

$$ \int e^{2x}\sin 2x \, dx = \frac{e^{2x}(2\sin 2x - 2\cos 2x)}{8} = \frac{e^{2x}(\sin 2x - \cos 2x)}{4} + C $$ $$ ye^{2x} = \frac{e^{2x}(\sin 2x - \cos 2x)}{4} + C $$

Using $$y(0) = 3/4$$:

$$ \frac{3}{4} = \frac{0 - 1}{4} + C = -\frac{1}{4} + C \Rightarrow C = 1 $$ $$ y = \frac{\sin 2x - \cos 2x}{4} + e^{-2x} $$

At $$x = \pi/8$$:

$$ \sin(\pi/4) = \frac{\sqrt{2}}{2}, \quad \cos(\pi/4) = \frac{\sqrt{2}}{2} $$ $$ y(\pi/8) = \frac{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{4} + e^{-\pi/4} = 0 + e^{-\pi/4} = e^{-\pi/4} $$

The correct answer is Option (3): $$e^{-\pi/4}$$.