Answer: roots \(x=2,-4,-4\), and \(y=1\).

(a) From

\(\log_2(y+1)=3-2\log_2x,\)

use \(3=\log_28\) and \(2\log_2x=\log_2x^2\):

\(\log_2(y+1)=\log_2\left(\frac8{x^2}\right).\)

Hence

\(y+1=\frac8{x^2},\)

so

\(x^2(y+1)=8.\)

From

\(\log_2(x+2)=2+\log_2y,\)

we get

\(\log_2(x+2)=\log_2(4y).\)

Therefore

\(x+2=4y.\)

So

\(y=\frac{x+2}{4}.\)

Substitute this into \(x^2(y+1)=8\):

\(x^2\left(\frac{x+2}{4}+1\right)=8.\)

Thus

\(x^2\left(\frac{x+6}{4}\right)=8.\)

Multiplying by \(4\),

\(x^2(x+6)=32.\)

So

\(x^3+6x^2-32=0.\)

(b) Test \(x=2\):

\(2^3+6(2^2)-32=8+24-32=0.\)

So \(x-2\) is a factor. Dividing gives

\(x^3+6x^2-32=(x-2)(x^2+8x+16).\)

Hence

\(x^3+6x^2-32=(x-2)(x+4)^2.\)

Therefore the roots are

\(x=2,\quad x=-4,\quad x=-4.\)

(c) The value \(x=-4\) is not valid in the original logarithmic equations, since \(\log_2x\) is not defined for a negative value of \(x\).

So \(x=2\). Using \(x+2=4y\),

\(2+2=4y.\)

Therefore

\(y=1.\)