Đáp án:

$\begin{array}{l}
a)A = 1 + 4 + {4^2} + {4^3} + ... + {4^{59}}\\
 = \left( {1 + 4} \right) + {4^2}\left( {1 + 4} \right) + ... + {4^{58}}\left( {1 + 4} \right)\\
 = 5 + {4^2}.5 + ... + {4^{58}}.5\\
 = \left( {1 + {4^2} + ... + {4^{58}}} \right).5 \vdots 5\\
b)A = 1 + 4 + {4^2} + {4^3} + ... + {4^{59}}\\
 = \left( {1 + 4 + {4^2}} \right) + \left( {{4^3} + {4^4} + {4^5}} \right) + ... + \left( {{4^{57}} + {4^{58}} + {4^{59}}} \right)\\
 = 21 + {4^3}.21 + ... + {4^{57}}.21\\
 = \left( {1 + {4^3} + ... + {4^{57}}} \right).21 \vdots 21\\
c)A \vdots 5;A \vdots 21\\
 \Leftrightarrow A \vdots \left( {5.21} \right)\\
 \Leftrightarrow A \vdots 105\\
d)A = 1 + 4 + {4^2} + {4^3} + ... + {4^{59}}\\
 \Leftrightarrow 4A = 4 + {4^2} + {4^3} + ... + {4^{60}}\\
 \Leftrightarrow 4A - A = 3A = {4^{60}} - 1\\
 \Leftrightarrow A = \dfrac{{{4^{60}} - 1}}{3}\\
e)3A + 1 = {4^x}\\
 \Leftrightarrow 3.\dfrac{{{4^{60}} - 1}}{3} + 1 = {4^x}\\
 \Leftrightarrow {4^{60}} - 1 + 1 = {4^x}\\
 \Leftrightarrow {4^{60}} = {4^x}\\
 \Leftrightarrow x = 60
\end{array}$