We need to evaluate $L=\lim_{x\to \frac\pi2}\frac{4\sqrt2(\sin 3x+\sin x)}{\left(2\sin 2x\sin \frac{3x}{2}+\cos \frac{5x}{2}\right)-\left(\sqrt2+\sqrt2\cos 2x+\cos \frac{3x}{2}\right)}.$

First simplify the numerator using $\sin 3x+\sin x=2\sin\frac{3x+x}{2}\cos\frac{3x-x}{2}=2\sin 2x\cos x.$ So numerator becomes $4\sqrt2(\sin 3x+\sin x)=4\sqrt2\cdot 2\sin 2x\cos x=8\sqrt2\sin 2x\cos x.$ Since $\sin 2x=2\sin x\cos x$, $8\sqrt2\sin 2x\cos x=16\sqrt2\sin x\cos^2 x.$

Now simplify the denominator: $D=2\sin 2x\sin\frac{3x}{2}+\cos\frac{5x}{2}-\sqrt2-\sqrt2\cos 2x-\cos\frac{3x}{2}.$ Use $2\sin A\sin B=\cos(A-B)-\cos(A+B).$ With $A=2x, B=\frac{3x}{2}$, $2\sin 2x\sin\frac{3x}{2}=\cos\frac{x}{2}-\cos\frac{7x}{2}.$ Hence $D=\cos\frac{x}{2}-\cos\frac{7x}{2}+\cos\frac{5x}{2}-\sqrt2-\sqrt2\cos 2x-\cos\frac{3x}{2}.$ Group cosine terms: $D=\left(\cos\frac{x}{2}-\cos\frac{3x}{2}\right)+\left(\cos\frac{5x}{2}-\cos\frac{7x}{2}\right)-\sqrt2(1+\cos 2x).$ Now use $\cos C-\cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}.$ Then $\cos\frac{x}{2}-\cos\frac{3x}{2}=2\sin x\sin\frac{x}{2},$ $\cos\frac{5x}{2}-\cos\frac{7x}{2}=2\sin 3x\sin\frac{x}{2}.$ So $D=2\sin\frac{x}{2}(\sin x+\sin 3x)-\sqrt2(1+\cos 2x).$ Again, $\sin x+\sin 3x=2\sin 2x\cos x.$ Therefore $D=4\sin\frac{x}{2}\sin 2x\cos x-\sqrt2(1+\cos 2x).$ Also, $1+\cos 2x=2\cos^2 x,$ so $D=4\sin\frac{x}{2}\sin 2x\cos x-2\sqrt2\cos^2 x.$ Using $\sin 2x=2\sin x\cos x$, $D=8\sin\frac{x}{2}\sin x\cos^2 x-2\sqrt2\cos^2 x.$ Factor $2\cos^2 x$: $D=2\cos^2 x\left(4\sin\frac{x}{2}\sin x-\sqrt2\right).$ Now use $\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}$: $4\sin\frac{x}{2}\sin x=8\sin^2\frac{x}{2}\cos\frac{x}{2}.$ So $D=2\cos^2 x\left(8\sin^2\frac{x}{2}\cos\frac{x}{2}-\sqrt2\right).$ At $x=\frac\pi2$, this bracket becomes $0$, so factor further. Let $t=\frac{x}{2}.$ Then $8\sin^2 t\cos t-\sqrt2=8(1-\cos^2 t)\cos t-\sqrt2.$ When $t\to \frac\pi4$, $\cos t\to \frac{1}{\sqrt2}$. Put $u=\cos t$: $8(1-u^2)u-\sqrt2=8u-8u^3-\sqrt2.$ Since $u=\frac1{\sqrt2}$ is a root, $8u-8u^3-\sqrt2=-(u-\tfrac1{\sqrt2})\cdot 8\left(u^2+\frac{u}{\sqrt2}-\frac12\right).$ A quicker factorization is obtained directly as $8\sin^2 t\cos t-\sqrt2=2\sqrt2(2\cos t+\sqrt2)(\cos t-\tfrac1{\sqrt2})^2.$ But near $t=\frac\pi4$, the simplest way is to use $\cos t-\frac1{\sqrt2}= -2\sin\frac{t+\pi/4}{2}\sin\frac{t-\pi/4}{2},$ though an even cleaner route is to use local behavior after factoring one $\cos x$. Instead, let us rewrite from a previous stage.

From numerator and denominator expressions: $N=16\sqrt2\sin x\cos^2 x,$ $D=2\cos^2 x\left(4\sin\frac{x}{2}\sin x-\sqrt2\right).$ Cancel $2\cos^2 x$: $L=\lim_{x\to \pi/2}\frac{8\sqrt2\sin x}{4\sin\frac{x}{2}\sin x-\sqrt2}.$ Now this is still $\frac{8}{0}$-type? Check carefully at $x=\pi/2$: $4\sin\frac\pi4\sin\frac\pi2=4\cdot \frac1{\sqrt2}\cdot 1=2\sqrt2,$ so denominator becomes $2\sqrt2-\sqrt2=\sqrt2\neq 0.$ Hence the limit is obtained by direct substitution.

Substitute $x=\frac\pi2$: