The value of the limit
`lim_(x to pi/2) (4sqrt2)(sin3x +sinx)/((2 sin 2x sin. (3x)/2 +cos. (5x)/2)-(sqrt(2)+sqrt(2)cos 2x +cos. (3x)/2))`
is _______

Answer

To solve the limit

\[ \lim_{x \to \frac{\pi}{2}} \frac{4\sqrt{2}(\sin 3x + \sin x)}{(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}) - (\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2})} \]

we will follow these steps:

### Step 1: Evaluate the numerator

First, we evaluate the numerator as \( x \to \frac{\pi}{2} \):

\[ \sin 3x + \sin x \to \sin(3 \cdot \frac{\pi}{2}) + \sin(\frac{\pi}{2}) = \sin(\frac{3\pi}{2}) + 1 = -1 + 1 = 0 \]

Thus, the numerator approaches \( 0 \).

### Step 2: Evaluate the denominator

Next, we evaluate the denominator:

\[ 2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2} - (\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}) \]

Calculating each term as \( x \to \frac{\pi}{2} \):

- \( \sin 2x \to \sin(\pi) = 0 \)
- \( \sin \frac{3x}{2} \to \sin(\frac{3\pi}{4}) = \frac{\sqrt{2}}{2} \)
- \( \cos \frac{5x}{2} \to \cos(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2} \)
- \( \cos 2x \to \cos(\pi) = -1 \)
- \( \cos \frac{3x}{2} \to \cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \)

Substituting these values into the denominator:

\[ 2(0)(\frac{\sqrt{2}}{2}) + (-\frac{\sqrt{2}}{2}) - (\sqrt{2} + \sqrt{2}(-1) - \frac{\sqrt{2}}{2}) \]
\[ = 0 - \frac{\sqrt{2}}{2} - (\sqrt{2} - \sqrt{2} + \frac{\sqrt{2}}{2}) = -\frac{\sqrt{2}}{2} - 0 = -\frac{\sqrt{2}}{2} \]

### Step 3: Form the limit

Now we have:

\[ \lim_{x \to \frac{\pi}{2}} \frac{4\sqrt{2} \cdot 0}{-\frac{\sqrt{2}}{2}} = \frac{0}{-\frac{\sqrt{2}}{2}} = 0 \]

### Conclusion

Thus, the value of the limit is:

\[ \boxed{0} \]