Answer: \(x=3,\ y=2\).

First write each equation using powers with the same base.

Since \(9=3^2\) and \(243=3^5\),

\(3^x\times 9^{y-1}=3^x\times(3^2)^{y-1} =3^{x+2y-2}.\)

Thus

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

so

\(x+2y=7.\)

For the second equation, write every term as a power of \(2\):

\(8\times2^{y-\frac12}=2^3\times2^{y-\frac12} =2^{y+\frac52}.\)

Also,

\(4\sqrt2=2^2\cdot2^{\frac12}=2^{\frac52}.\)

So

\(\frac{2^{2x+1}}{4\sqrt2} =\frac{2^{2x+1}}{2^{\frac52}} =2^{2x-\frac32}.\)

Equating powers gives

\(y+\frac52=2x-\frac32.\)

Hence

\(2x-y=4.\)

Now solve

\(x+2y=7,\qquad 2x-y=4.\)

From \(2x-y=4\), \(y=2x-4\). Substitute into \(x+2y=7\):

\(x+2(2x-4)=7.\)

So

\(5x=15,\)

and \(x=3\). Then

\(y=2(3)-4=2.\)