Answer: (i) \(x=1.19\). (ii) \(y=\dfrac73\).

(i) Take logarithms:

\((3x-1)\log2=\log6.\)

So

\(3x-1=\frac{\log6}{\log2}.\)

Hence

\(x=\frac{\frac{\log6}{\log2}+1}{3} =1.19\)

to 3 significant figures.

(ii) Use

\(1=\log_3 3 \quad\text{and}\quad \frac{1}{\log_y3}=\log_3y.\)

Then

\(\log_3(y+14)=\log_33+2\log_3y.\)

Therefore

\(\log_3(y+14)=\log_3(3y^2).\)

So

\(y+14=3y^2.\)

This gives

\(3y^2-y-14=0.\)

Factorise:

\((3y-7)(y+2)=0.\)

The logarithms require \(y\gt 0\), so \(y=-2\) is not valid. Hence

\(\boxed{y=\frac73}.\)