Answer: (a)(ii) \(y=-3+\sqrt6\); (b) \(2+3\log_a9\).

(a)(i) For the logarithms to be defined, both arguments must be positive:

\(x+3\gt 0 \quad\text{and}\quad 2x-1\gt 0.\)

The second condition gives

\(x\gt \frac12.\)

This is stronger than \(x\gt -3\), so \(x\) must be greater than \(\frac12\).

(a)(ii) The equation is

\(\frac{\log_a6}{\log_a(y+3)}=2.\)

Using change of base,

\(\frac{\log_a6}{\log_a(y+3)}=\log_{y+3}6.\)

So

\(\log_{y+3}6=2.\)

Therefore

\((y+3)^2=6.\)

Since \(y+3\) is a positive logarithm base,

\(y+3=\sqrt6.\)

Hence

\(y=-3+\sqrt6.\)

(b) Let \(L=\log_a9\) and \(B=\log_ab\).

The expression is

\(L+B\log_{\sqrt b}(9a).\)

By change of base,

Now

\(\log_a(9a)=\log_a9+\log_aa=L+1\)

and

\(\log_a\sqrt b=\log_ab^{\frac12}=\frac12B.\)

Therefore

\(L+B\left(\frac{L+1}{\frac12B}\right) =L+2(L+1).\)

So the expression is

\(2+3\log_a9.\)