$\text{ In the isosceles triangle }\underset{\_}{\mathrm{ABC},}\left|\overrightarrow{AB}\right|=\left|\overrightarrow{BC}\right|=8,\text{ a point }\mathrm{E}\text{ divides }\mathrm{AB}\text{ internally in }$$\text{ the ratio }1:3,\text{ then the cosine of the angle between }\overrightarrow{CE}\text{ and }\overrightarrow{CA}\text{ is }:\text{ (where }\left|\overrightarrow{CA}\right|=12\text{ ) }$

$\text{ Given }\left|\overrightarrow{b}\right|=|\overrightarrow{b}-\overrightarrow{c}|=8\text{ and }\left|\overrightarrow{c}\right|=12$

$$\begin{gathered} |\overrightarrow{b}{|}^2=|\overrightarrow{b}-\overrightarrow{c}{|}^2 \\ |\overrightarrow{b}{|}^2={\left|\overrightarrow{b}\right|}^2+{\left|\overrightarrow{c}\right|}^2-2\left|\overrightarrow{b}·\overrightarrow{c}\right| \\ \overrightarrow{b}·\overrightarrow{c}=\frac{144}{2}=72 \end{gathered}$$

$$\begin{gathered} {\left(\overrightarrow{c}-\frac{\overrightarrow{b}}{4}\right)}^2={\left|\overrightarrow{c}\right|}^2-2\overrightarrow{c}·\frac{\overrightarrow{b}}{4}+\frac{{\overrightarrow{b}}^2}{16} \\ =144-\frac{72}{4}+\frac{64}{16}=112 \\ \left|\overrightarrow{c}-\frac{\overrightarrow{b}}{4}\right|=\sqrt{112} \end{gathered}$$

$\overrightarrow{CE}=\overrightarrow{OE}-\overrightarrow{OC}=\frac{\overrightarrow{b}}{4}-\overrightarrow{c}$ $\overrightarrow{CA}=-\overrightarrow{c}$

$\cos \theta =\frac{\overrightarrow{c}\left(\overrightarrow{c}-\frac{\overrightarrow{b}}{4}\right)}{\left|\overrightarrow{c}\right|\left|\overrightarrow{c}-\frac{\overrightarrow{b}}{4}\right|}=\frac{\overrightarrow{c^2}-\frac{\overrightarrow{c}\cdot \overrightarrow{b}}{4}}{12\left|\overrightarrow{c}-\frac{\overrightarrow{b}}{4}\right|}$

$=\frac{144-18}{12\sqrt{112}}=\frac{126}{12\times 4\sqrt{7}}$

$=\frac{3\sqrt{7}}{8}$