8. In $\triangle A B C$ and $\triangle A E F$, $B$ is the midpoint of $E F$, $A B=E F=1, B C=6, C A=\sqrt{33}$. If $A B \cdot A E + A C \cdot A F=2$, then the cosine value of the angle between $E F$ and $B C$ is $\qquad$
8. $\frac{2}{3}$.

Since $A B \cdot A E+A C \cdot A F=2$, we have $A B \cdot(A B+B E)+A C \cdot(A B+B F)=2$, which means
$$
A B^{2}+A B \cdot B E+A C \cdot A B+A C \cdot B F=2 \text {. }
$$

Since $A B^{2}=1, B E=-B F$,
$$
\boldsymbol{A} \cdot \boldsymbol{A B}=\sqrt{33} \times 1 \times \frac{33+1-36}{2 \times \sqrt{33} \times 1}=-1 \text {, }
$$

Therefore, $1+B F \cdot(A C-A B)-1=2$, which means
$$
B F \cdot B C=2 \text {. }
$$

Let the angle between $\boldsymbol{E F}$ and $\boldsymbol{B} \boldsymbol{C}$ be $\theta$, then
$$
|\boldsymbol{B F}| \cdot|\boldsymbol{B} \boldsymbol{C}| \cos \theta=2 \Rightarrow 3 \cos \theta=2 \Rightarrow \cos \theta=\frac{2}{3} .
$$