(1) If $4^{a}-3 a^{b}=16, \log _{2} a=\frac{a+1}{b}$, then $a^{b}=$
(1) 16 Hint: From $\log _{2} a=\frac{a+1}{b}$, we get $a^{b}=2^{a+1}$. Substituting into $4^{a}-3 a^{b}=16$ yields $2^{2 a}-6 \times 2^{a}-16=0$. Solving this, we get: $2^{a}=8, 2^{a}=-2$ (discard), so, $a^{b}=$ $2^{a+1}=16$.