5. (1990 American High School Mathematics Examination) First, select $a$ from $\{1,2,3, \cdots, 99,100\}$, then select $b$ from the same set. The probability that the last digit of $3^{a}+7^{b}$ is 8 is ( ).
A. $\frac{1}{16}$
B. $\frac{1}{8}$
C. $\frac{3}{16}$
D. $\frac{1}{5}$
E. $\frac{1}{4}$
5. C. Reason: The last digit of $3^{n}$ changes periodically with the increase of $n$ (starting from 1) in the sequence $3,9,7,1$, and the last digit of $7^{m}$ changes periodically in the sequence $7,9,3,1$.

Since their periods are both 4, and among the 100 numbers from 1 to 100, the numbers of the form $4 k, 4 k+1, 4 k+2, 4 k+3$ are evenly distributed (each has 25 numbers),
In the above last digits, only the cases $7+1, 1+7, 9+9$ result in a last digit of 8. That is,
\begin{tabular}{c|c|c|c}
\hline$a$ & $4 k+2$ & $4 k+3$ & $4 k$ \\
\hline$b$ & $4 l+2$ & $4 l$ & $4 l+1$ \\
\hline The last digit of $3^{a}+7^{b}$ & 8 & 8 & 8 \\
\hline
\end{tabular}

In this way, the number of pairs $(a, b)$ chosen from the four types of numbers is $C_{4}^{1} C_{4}^{4}=16$, and only 3 of them meet the problem's requirements. Therefore, the required probability (possibility) is $\frac{3}{16}$.