Answer: (a)(i) \(7\), (ii) \(\dfrac17\). (b) \(y=\dfrac13\). (c) \(x=\pm\sqrt3\).

(a)(i) Since \(a^7=b\), the power to which \(a\) must be raised to give \(b\) is \(7\). Therefore

\(\log_a b=7.\)

(a)(ii) Using the reciprocal rule for logarithms,

\(\log_b a=\frac{1}{\log_a b}=\frac17.\)

(b) Convert from logarithmic to exponential form:

\(y=81^{-1/4}.\)

Since \(81=3^4\),

\(y=(3^4)^{-1/4}=3^{-1}=\frac13.\)

(c) Write everything as powers of \(2\):

\(\frac{32^{x^2-1}}{4^{x^2}} = \frac{(2^5)^{x^2-1}}{(2^2)^{x^2}} = 2^{5(x^2-1)-2x^2} = 2^{3x^2-5}.\)

Also \(16=2^4\), so

\(2^{3x^2-5}=2^4.\)

Hence the exponents are equal:

\(3x^2-5=4.\)

So

\(3x^2=9 \quad\Rightarrow\quad x^2=3.\)

Therefore

\(\boxed{x=\pm\sqrt3}.\)