1. **State the problem:** We are given that $x$ and $y$ are positive real numbers and satisfy the system:\n$$\log_5(x+y) + \log_5(x-y) = 3$$\nand\n$$\log_2 y - \log_2 x = 1 - \log_2 3.$$\nWe want to find the value of $xy$.\n\n2. **Use the properties of logarithms:** The sum of logarithms with the same base can be rewritten as the logarithm of a product. So:\n$$\log_5(x+y) + \log_5(x-y) = \log_5((x+y)(x-y)) = 3.$$\n\n3. **Rewrite the equation:** This means:\n$$ (x+y)(x-y) = 5^3 = 125.$$\n\n4. **Expand:** We get\n$$x^2 - y^2 = 125.$$\n\n5. **For the second equation:** Using log subtraction rule:\n$$\log_2 y - \log_2 x = \log_2 \frac{y}{x} = 1 - \log_2 3.$$\nThis implies\n$$\log_2 \frac{y}{x} = \log_2 2 - \log_2 3 = \log_2 \frac{2}{3},$$\nso\n$$\frac{y}{x} = \frac{2}{3}.$$\n\n6. **Express $y$:** From above,\n$$y = \frac{2}{3} x.$$\n\n7. **Substitute $y$ into the expanded equation:**\n$$x^2 - \left(\frac{2}{3} x\right)^2 = 125,$$\nwhich simplifies to\n$$x^2 - \frac{4}{9} x^2 = 125.$$\n\n8. **Combine like terms:**\n$$x^2 \left(1 - \frac{4}{9}\right) = 125,$$\nwhich is\n$$x^2 \times \frac{5}{9} = 125.$$\n\n9. **Solve for $x^2$:**\n$$x^2 = 125 \times \frac{9}{5} = 125 \times 1.8 = 225.$$\n\n10. **Find $x$:** Since $x$ is positive,\n$$x = 15.$$\n\n11. **Find $y$:**\n$$y = \frac{2}{3} \times 15 = 10.$$\n\n12. **Calculate $xy$:**\n$$xy = 15 \times 10 = 150.$$\n\n**Final answer:** $xy = 150$.