8. B. As shown in Figure 2, given the hyperbola $y=\frac{2}{x}(x>0)$ intersects the sides $C B$ and $B A$ of rectangle $O A B C$ at points $E$ and $F$ respectively, and $A F=B F$, connect $E F$. Then the area of $\triangle O E F$ is $\qquad$
8. B. $\frac{3}{2}$.

Let point $B(a, b)$. Then point $F\left(a, \frac{b}{2}\right) 、 E\left(\frac{2}{b}, b\right)$.
Since point $F$ is on the hyperbola $y=\frac{2}{x}$, we have $a b=4$.
Therefore, $S_{\triangle O E F}=S_{\text {rectangle } OFBC }-S_{\triangle O E C}-S_{\triangle F B E}$ $=\frac{1}{2}\left(\frac{b}{2}+b\right) a-\frac{1}{2} b \times \frac{2}{b}-\frac{1}{2} \times \frac{b}{2}\left(a-\frac{2}{b}\right)$ $=\frac{1}{2}(a b+1-2)=\frac{3}{2}$.