The area, enclosed by the curves $$y = \sin x + \cos x$$ and $$y = |\cos x - \sin x|$$ and the lines $$x = 0$$, $$x = \frac{\pi}{2}$$, is:

Given curves,

$$y=\sin x+\cos x$$

and

$$y=|\cos x-\sin x|$$

for

$$0\le x\le\frac\pi2$$

Now,

$$\cos x-\sin x\ge0$$

for

$$0\le x\le\frac\pi4$$

Hence,

$$|\cos x-\sin x|=\cos x-\sin x$$

Also,

$$\cos x-\sin x<0$$

for

$$\frac\pi4\le x\le\frac\pi2$$

Hence,

$$|\cos x-\sin x|=\sin x-\cos x$$

Therefore, area is

$$\int_0^{\pi/4}\left[(\sin x+\cos x)-(\cos x-\sin x)\right]dx$$

$$+\int_{\pi/4}^{\pi/2}\left[(\sin x+\cos x)-(\sin x-\cos x)\right]dx$$

Simplifying,

$$=\int_0^{\pi/4}2\sin x\,dx+\int_{\pi/4}^{\pi/2}2\cos x\,dx$$

Now,

$$\int_0^{\pi/4}2\sin x\,dx=2[-\cos x]_0^{\pi/4}$$

$$=2\left(1-\frac1{\sqrt2}\right)$$

Also,

$$\int_{\pi/4}^{\pi/2}2\cos x\,dx=2[\sin x]_{\pi/4}^{\pi/2}$$

$$=2\left(1-\frac1{\sqrt2}\right)$$

Hence total area is

$$4\left(1-\frac1{\sqrt2}\right)$$

$$=4-2\sqrt2$$

Therefore, the required area is

$$\boxed{4-2\sqrt2}$$