Answer: (a) \(x=64\). (b) \((x,y)=(1,0)\) or \(\left(\frac12,\frac14\right)\).

(a) Let \(u=x^{\frac16}\). Then \(x^{\frac13}=u^2\), so

\(u^2-u=2.\)

Thus

\(u^2-u-2=0,\)

and

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

Since \(u=x^{\frac16}\geqslant0\), reject \(u=-1\). Hence \(u=2\), and

\(x=2^6=64.\)

(b) From \(\lg(x+2y)=0\),

\(x+2y=1.\)

So \(x=1-2y\). Substitute this into \(x^2+4xy+y=1\):

\((1-2y)^2+4(1-2y)y+y=1.\)

Expanding and simplifying gives

\(y-4y^2=0.\)

So

\(y(1-4y)=0.\)

Hence \(y=0\) or \(y=\frac14\). Using \(x=1-2y\),

\((x,y)=(1,0)\quad\text{or}\quad\left(\frac12,\frac14\right).\)