$|\overset{\large\rightharpoonup}{AB}| = \sqrt{(-4)^2 + 3^2} = 5$，$|\overset{\large\rightharpoonup}{AC}| = \sqrt{(\frac{24}{5})^2 + (\frac{7}{5})^2} = \sqrt{\frac{576+49}{25}} = 5$。 $(1)$ $\overset{\large\rightharpoonup}{BC} = \overset{\large\rightharpoonup}{AC} - \overset{\large\rightharpoonup}{AB} = (\frac{44}{5}, -\frac{8}{5})$，$|\overset{\large\rightharpoonup}{BC}| = \sqrt{(\frac{44}{5})^2 + (\frac{-8}{5})^2} = \sqrt{80} = 4\sqrt{5} \neq 5$。 $(2)$ $|\overset{\large\rightharpoonup}{AB}| = |\overset{\large\rightharpoonup}{AC}| = 5$，故 $\Delta ABC$ 為等腰三角形。（正確） $(3)$ 面積 $= \dfrac{1}{2} |(-4)(\frac{7}{5}) - 3(\frac{24}{5})| = \dfrac{1}{2} |-\frac{28}{5} - \frac{72}{5}| = \dfrac{1}{2} |-20| = 10$。（正確） $(4)$ 因 $\overline{AB} = \overline{AC}$，故 $\angle B = \angle C \implies \sin B = \sin C$。 $(5)$ $\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC} = -4(\frac{24}{5}) + 3(\frac{7}{5}) = -\frac{75}{5} = -15$。$\cos A = \dfrac{-15}{5 \times 5} = -\dfrac{3}{5} < 0$，而 $\angle B$ 為銳角，$\cos B > 0$，故 $\cos A < \cos B$。 故選 $(2)(3)$。