Answer: \(xy=256\); \(x=-1\) or \(x=0\).

(a) Since

\(\log_4y=\frac12\log_2y,\)

we have

\(2\log_4y=\log_2y.\)

Therefore

\(\log_2x+2\log_4y =\log_2x+\log_2y.\)

So

\(\log_2(xy)=8.\)

Hence

\(xy=2^8=256.\)

(b) Let

\(y=2^x.\)

Then \(2^{2x}=y^2\), so

\(2^{2x+1}=2y^2,\qquad 2^{x+1}=2y,\qquad 2^x=y.\)

The equation becomes

\(2y^2-2y-y+1=0.\)

So

\(2y^2-3y+1=0.\)

Factorising,

\((2y-1)(y-1)=0.\)

Hence

\(y=\frac12\quad\text{or}\quad y=1.\)

Since \(y=2^x\),

\(2^x=\frac12\quad\text{or}\quad 2^x=1.\)

Therefore

\(x=-1\quad\text{or}\quad x=0.\)