Answer: (a) \(x=256\). (b) \(x=\frac{\ln5-1}{2}\).

Apply the laws of logarithms carefully, paying attention to the base of each logarithm and any restrictions on the argument.

For part (a), use base \(2\):

\(\log_2x^2+\log_{16}x=18\).

Since \(\log_2x^2=2\log_2x\) and \(\log_{16}x=\frac14\log_2x\),

\(\frac94\log_2x=18\).

So \(\log_2x=8\), giving \(x=2^8=256\).

For part (b), let \(u=\mathrm e^{2x+1}\). Then \(\mathrm e^{-2x-1}=\frac1u\), so

\(u-\frac{10}{u}=3\).

Multiplying by \(u\) gives \(u^2-3u-10=0\).

Thus \((u-5)(u+2)=0\). Since \(u\gt0\), \(u=5\).

Therefore \(\mathrm e^{2x+1}=5\), so \(2x+1=\ln5\), and

\(x=\frac{\ln5-1}{2}\).

This gives the required answer: (a) \(x=256\). (b) \(x=\frac{\ln5-1}{2}\).