Answer: \(\lg\left(\frac{1000a^2}{b^4}\right)\); \(a=2\) or \(a=64\).

The key point is to rewrite the equation using logarithm laws or a suitable substitution.

(a) Since \(3=\lg1000\),

\(3+2\lg a-4\lg b=\lg1000+\lg a^2-\lg b^4.\)

Using the laws of logarithms,

(b) Let \(u=\log_4a\). Then \(\log_a4=\frac1u\). The equation becomes

\(\frac3u+2u=7.\)

Multiplying by \(u\),

\(2u^2-7u+3=0.\)

Factorising,

\((2u-1)(u-3)=0.\)

So

\(u=\frac12\quad\text{or}\quad u=3.\)

Therefore

\(\log_4a=\frac12\quad\text{or}\quad \log_4a=3.\)

Hence \(a=4^{1/2}=2\) or \(a=4^3=64\).