Given:

$$
\(\frac{1}{2}<x<\frac{3}{4}, \quad \frac{2}{7}<y<\frac{4}{7}\)
$$

we need to find which among the options can be a possible value of $x y$.
Step 1: Find the range for $x y$
- Minimum possible $x y=$ minimum $x \times$ minimum $y$, since both positive and increasing,

$$
\(x y_{\min }=\left(\frac{1}{2}\right) \times\left(\frac{2}{7}\right)=\frac{1}{7} \approx 0.1429\) .
$$

- Maximum possible $x y=\operatorname{maximum} x \times \operatorname{maximum} y$,

$$
\(x y_{\max }=\left(\frac{3}{4}\right) \times\left(\frac{4}{7}\right)=\frac{12}{28}=\frac{3}{7} \approx 0.4286\)
$$


Step 2: Check which options lie between 0.1429 and 0.4286
Given options:
a) $\(\frac{1}{7}=0.1429\)$ (equal to the minimum; since $x, y$ both strictly greater than given values, cannot equal minimum)
b) $\(\frac{2}{7}=0.2857\)$ (between min and max)
c) $\(\frac{3}{7}=0.4286\)$ (equal to max; same as minimum, can't equal max if strict inequality)
d) $\(\frac{4}{7}=0.5714\)$ (greater than $\(\max\)$; no)
e) $\(\frac{5}{7}=0.7143\)$ (greater than $\(\max\)$; no)

Step 3: Consider strict inequalities <
- $x y$ cannot be equal to min or max values, so $\(\frac{1}{7}\)$ or $\(\frac{3}{7}\)$ are not possible values exactly.
- The only possible choice within range and not equal to endpoints is $\(\frac{2}{7}\)$.

Final answer:
B. $\(\frac{2}{7}\)$ is a possible value of $x y$.