Đáp án:
$\begin{array}{l}
1){4^{x + 1}} - {3.2^{x + 2}} = 16\\
\Rightarrow {4^x}.4 - {3.2^x}{.2^2} - 16 = 0\\
\Rightarrow {4.2^{2x}} - {12.2^x} - 16 = 0\\
\Rightarrow {\left( {{2^x}} \right)^2} - {3.2^x} - 4 = 0\\
\Rightarrow \left[ \begin{array}{l}
{2^x} = 4\\
{2^x} = - 1\left( {loại} \right)
\end{array} \right. \Rightarrow x = 2
\end{array}$
$\begin{array}{l}
2)\\
{5.8^{x - 1}} - {3.2^{5 - 3x}} + 7 = 0\\
\Rightarrow \frac{5}{8}{.2^{3x}} - {3.2^5}.\frac{1}{{{2^{3x}}}} + 7 = 0\\
\Rightarrow \frac{5}{8}{.2^{3x}} - 96.\frac{1}{{{2^{3x}}}} + 7 = 0\\
\Rightarrow \frac{5}{8}a - \frac{{96}}{a} + 7 = 0\left( {a = {2^{3x}} > 0} \right)\\
\Rightarrow \frac{5}{8}{a^2} + 7a - 96 = 0\\
\Rightarrow a = 8\\
\Rightarrow {2^{3x}} = 8\\
\Rightarrow 3x = 3\\
\Rightarrow x = 1
\end{array}$
