2. $\frac{\sqrt{15}}{15}$.

Notice that $\overrightarrow{A B} \cdot \overrightarrow{B C}=c a \cos (\pi-B)=-\frac{a^{2}+c^{2}-b^{2}}{2}$, then the condition is equivalent to
$$
\begin{array}{l}
c^{2}+a^{2}-b^{2}=2\left(a^{2}+b^{2}-c^{2}\right)=4\left(b^{2}+c^{2}-a^{2}\right) \\
\Rightarrow a^{2}: b^{2}: c^{2}=6: 3: 5 \\
\Rightarrow \cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3+5-6}{2 \cdot \sqrt{3} \cdot \sqrt{5}}=\frac{\sqrt{15}}{15}
\end{array}
$$
2. Given in $\triangle A B C$, $\overrightarrow{A B} \cdot \overrightarrow{B C}=2 \overrightarrow{B C} \cdot \overrightarrow{C A}=4 \overrightarrow{C A} \cdot \overrightarrow{A B}$, then the cosine value of the largest angle in $\triangle A B C$ is