$8.13 \frac{x-7}{\sqrt{4 x^{2}-19 x+12}}<0$.

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$8.13 \frac{x-7}{\sqrt{4 x^{2}-19 x+12}}<0$.
8.13 The given inequality can only be considered under the condition $4 x^{2}-19 x+12>0$, from which we find that $x<0.75$ or $x>4$.

For these values of $x$, we have $\sqrt{4 x^{2}-19 x+12}>0$ and it is sufficient to solve the inequality $x-7<0$, i.e.:

$$
x<7
$$

From the system of inequalities (1), (2) we obtain that $-\infty<x<0.75$ or $4<x<7$.

Answer: $(-\infty ; 0.75) \cup(4,7)$.