Đáp án:

 1) 07; 2) 76

Giải thích các bước giải:

1) $\begin{array}{l}
A = {7^{{9^{{7^9}}}}}\\
 + ){7^4} \equiv 1\left( {\bmod 100} \right)\\
 + ){9^2} \equiv 1\left( {\bmod 4} \right)\\
 + )7 \equiv 1\left( {\bmod 2} \right) \Rightarrow {7^9} \equiv 1\left( {\bmod 2} \right) \Rightarrow {7^9} = 2k + 1\left( {k \in N} \right)\\
 \Rightarrow {9^{{7^9}}} = {9^{2k + 1}} = {\left( {{9^2}} \right)^k}.9 \equiv {1^k}.9\left( {\bmod 4} \right) \Rightarrow {9^{{7^9}}} \equiv 1\left( {\bmod 4} \right)\\
 \Rightarrow {9^{{7^9}}} = 4q + 1\left( {q \in N} \right)\\
 \Rightarrow {7^{{9^{{7^9}}}}} = {7^{4q + 1}} = {\left( {{7^4}} \right)^q}.7 \equiv {1^q}.7\left( {\bmod 100} \right)\\
 \Rightarrow {7^{{9^{{7^9}}}}} \equiv 7\left( {\bmod 100} \right)
\end{array}$

Vậy 2 chữ số tận cùng cần tìm là 07.

2) $\begin{array}{l}
{1978^{{{1986}^8}}}\\
 + )1978 \equiv 0\left( {\bmod 2} \right) \Rightarrow {1978^2} \equiv 0\left( {\bmod 4} \right)\\
 \Rightarrow {1978^{{{1986}^8}}} \equiv 0\left( {\bmod 4} \right)\left( 1 \right)\\
 + )1978 \equiv 3\left( {\bmod 5} \right) \Rightarrow {1978^{{{1986}^8}}} \equiv {3^{{{1986}^8}}}\left( {\bmod 5} \right)\left( 2 \right)\\
{3^4} \equiv 1\left( {\bmod 5} \right)\\
1986 \equiv 0\left( {\bmod 2} \right) \Rightarrow {1986^2} \equiv 0\left( {\bmod 4} \right) \Rightarrow {1986^8} \equiv 0\left( {\bmod 4} \right)\\
 \Rightarrow {1986^8} = 4k\left( {k \in N} \right)\\
 \Rightarrow {3^{{{1986}^8}}} = {3^{4k}} = {\left( {{3^4}} \right)^k} \equiv 1\left( {\bmod 5} \right)\left( 3 \right)\\
\left( 2 \right),\left( 3 \right) \Rightarrow {1978^{{{1986}^8}}} \equiv 1\left( {\bmod 5} \right) \Rightarrow {1978^{{{1986}^8}}} \equiv 1\left( {\bmod 25} \right)\left( 4 \right)\\
\left( 1 \right),\left( 4 \right) \Rightarrow \left\{ \begin{array}{l}
{1978^{{{1986}^8}}} \equiv 0\left( {\bmod 4} \right)\\
{1978^{{{1986}^8}}} \equiv 1\left( {\bmod 25} \right)
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{1978^{{{1986}^8}}} \equiv 76\left( {\bmod 4} \right)\\
{1978^{{{1986}^8}}} \equiv 76\left( {\bmod 25} \right)
\end{array} \right.\\
 \Rightarrow {1978^{{{1986}^8}}} \equiv 76\left( {\bmod 100} \right)
\end{array}$

Vậy 2 chữ số tận cùng là 76.