Answer: \(x=-\frac12\), \(y=\frac14\).

First write each equation using a single base.

Since \(49=7^2\),

\(7^x\times49^y=7^x\times(7^2)^y=7^{x+2y}.\)

So

\(x+2y=0.\)

Also, \(125=5^3\) and \(\frac1{25}=5^{-2}\). Hence

\(5^{5x}\times125^{\frac23y}=5^{5x}\times(5^3)^{\frac23y}=5^{5x+2y}.\)

So

\(5x+2y=-2.\)

Now solve the simultaneous equations

\(x+2y=0,\qquad 5x+2y=-2.\)

Subtracting the first equation from the second gives

\(4x=-2,\)

so

\(x=-\frac12.\)

Then \(x+2y=0\) gives

\(-\frac12+2y=0,\)

so

\(y=\frac14.\)

Therefore

\(\boxed{x=-\frac12,\quad y=\frac14}.\)