Answer: \(x=2\), \(y=7\).

From

\(\log_3(x+y)=2,\)

we get

\(x+y=3^2=9.\)

From

\(2\log_3(x+1)=\log_3(y+2),\)

we get

\(\log_3((x+1)^2)=\log_3(y+2).\)

So

\((x+1)^2=y+2.\)

Since \(y=9-x\), substitute into this equation:

\((x+1)^2=11-x.\)

Expand and simplify:

\(x^2+2x+1=11-x,\)

so

\(x^2+3x-10=0.\)

Factorise:

\((x+5)(x-2)=0.\)

Thus

\(x=-5\quad\text{or}\quad x=2.\)

The value \(x=-5\) is invalid because \(x+1\) would be negative inside the logarithm.

Therefore

\(x=2.\)

Then

\(y=9-2=7.\)