Answer: \(x=28.7\); \(y=4\) or \(y=5\).

(a) Take logarithms of both sides of

\(5^{x-2}=3\times2^{2x+3}.\)

This gives

\((x-2)\log5=\log3+(2x+3)\log2.\)

Expand and collect the \(x\)-terms:

\(x\log5-2\log5=\log3+2x\log2+3\log2.\)

So

\(x(\log5-2\log2)=\log3+3\log2+2\log5.\)

Hence

\(x=\frac{\log3+3\log2+2\log5}{\log5-2\log2}.\)

Therefore

\(x=28.7\)

to 1 decimal place.

(b) Since \(2=\log_3 9\),

So the equation becomes

\(\log_3\left(\frac{y^2+11}{9}\right)=\log_3(y-1).\)

Therefore

\(\frac{y^2+11}{9}=y-1.\)

Multiply by \(9\):

\(y^2+11=9y-9.\)

Hence

\(y^2-9y+20=0.\)

Factorise:

\((y-4)(y-5)=0.\)

Both values satisfy \(y-1\gt 0\), so

\(y=4\quad\text{or}\quad y=5.\)