3log₃ y - 5log₂ x = -11; 4log₁/₂ x + log₃ y = -13.

Let $$a = \log_3 y$$ and $$b = \log_2 x$$. The system becomes:
1) $$3a - 5b = -11$$
2) $$4\log_{1/2} x + a = -13$$
Since $$\log_{1/2} x = \frac{\log_2 x}{\log_2 (1/2)} = \frac{b}{-1} = -b$$, the second equation is $$4(-b) + a = -13$$, which simplifies to $$a - 4b = -13$$.
Now we have a system of linear equations:
1) $$3a - 5b = -11$$
2) $$a - 4b = -13$$
From (2), $$a = 4b - 13$$. Substitute this into (1):
$$3(4b - 13) - 5b = -11$$
$$12b - 39 - 5b = -11$$
$$7b = 28$$
$$b = 4$$
Substitute $$b=4$$ back into $$a = 4b - 13$$:
$$a = 4(4) - 13 = 16 - 13 = 3$$
So, $$\log_3 y = 3$$ and $$\log_2 x = 4$$.
This means $$y = 3^3 = 27$$ and $$x = 2^4 = 16$$.
The solution is $$x=16, y=27$$.