Answer: (a)(i) \(1+p+2q\), (ii) \(3p-q-1\), (iii) \(\frac1p+\frac1q\). (b) \(x=\log_3\frac43\approx0.262\).

(a)(i)

\(\log_a(axy^2)=\log_a a+\log_a x+\log_a y^2.\)

So

\(\log_a(axy^2)=1+p+2q.\)

(a)(ii)

\(\log_a\left(\frac{x^3}{ay}\right)=\log_a x^3-\log_a a-\log_a y.\)

Therefore

\(\log_a\left(\frac{x^3}{ay}\right)=3p-1-q.\)

(a)(iii) By the reciprocal law,

\(\log_x a=\frac1{\log_a x}=\frac1p,\qquad \log_y a=\frac1{\log_a y}=\frac1q.\)

Hence

\(\log_x a+\log_y a=\frac1p+\frac1q.\)

(b) Let

\(m=3^x.\)

Then

\(3^{1+2x}=3(3^x)^2=3m^2.\)

The equation becomes

\(m-3m^2+4=0.\)

So

\(3m^2-m-4=0.\)

Factorise:

\((3m-4)(m+1)=0.\)

Since \(m=3^x\gt 0\), we take

\(m=\frac43.\)

Therefore

\(3^x=\frac43,\)

so

\(\boxed{x=\log_3\frac43\approx0.262}.\)