Answer: (a) \(x=-\frac17\). (b) \(b=a^2\) or \(b=a^{-1}\).

(a) Write all terms as powers of \(3\):

\(9^{5x}=(3^2)^{5x}=3^{10x},\)

\(27^{x-2}=(3^3)^{x-2}=3^{3x-6},\)

and

\(243=3^5.\)

So

\(\frac{3^{10x}}{3^{3x-6}}=3^5.\)

This gives

\(3^{10x-(3x-6)}=3^5,\)

so

\(3^{7x+6}=3^5.\)

Equating powers,

\(7x+6=5.\)

Hence

\(\boxed{x=-\frac17}.\)

(b) Let

\(t=\log_a b.\)

Then

\(\log_a\sqrt b=\frac12\log_a b=\frac12t,\)

and

\(\log_b a=\frac{1}{\log_a b}=\frac1t.\)

The equation becomes

\(\frac12t-\frac12=\frac1t.\)

Multiply by \(2t\):

\(t^2-t=2.\)

So

\(t^2-t-2=0.\)

Factorise:

\((t-2)(t+1)=0.\)

Thus \(t=2\) or \(t=-1\). Since \(t=\log_a b\),

\(\log_a b=2\quad\text{or}\quad \log_a b=-1.\)

Therefore

\(\boxed{b=a^2\text{ or }b=a^{-1}}.\)