Datos del problema:
$$ x+y=90^\circ \Rightarrow y=90^\circ-x, \qquad x+z=180^\circ \Rightarrow z=180^\circ-x. $$

Parte 1: Primer cociente
$$ x+y-z=90^\circ-(180^\circ-x)=x-90^\circ, \quad \sen(x-90^\circ)=-\cos x. $$
$$ x-y+z=x-(90^\circ-x)+(180^\circ-x)=x+90^\circ, \quad \sen(x+90^\circ)=\cos x. $$
Entonces:
$$ \frac{\sen(x+y-z)}{\sen(x-y+z)}=\frac{-\cos x}{\cos x}=-1. $$

Parte 2: Segundo cociente

$$ 2x+3z=2x+3(180^\circ-x)=540^\circ-x \Rightarrow \cos(540^\circ-x)=\cos(180^\circ+(360^\circ-x))=-\cos x. $$

$$ 4y-3z=4(90^\circ-x)-3(180^\circ-x)= -180^\circ-x. $$
$$ \sin(-180^\circ-x)=\sin x \Rightarrow \csc(4y-3z)=\frac{1}{\sin x}. $$

$$ x+y+z=90^\circ+(180^\circ-x)=270^\circ-x \Rightarrow \tan(270^\circ-x)=\tan(90^\circ-x)=\cot x. $$

Luego:
$$ \frac{\cos(2x+3z)\csc(4y-3z)}{\tan(x+y+z)} = \frac{(-\cos x)\left(\frac{1}{\sin x}\right)}{\cot x} = \frac{-\cot x}{\cot x}=-1. $$

Conclusión:
$$ Q=(-1)-(-1)=0. $$

Resultado final: $\boxed{Q=0}$.