Let $X$=number of people who carry the defective gene that causes inherited colon cancer.
By the report of the article in the Los Angeles Times (Dec. 3, 1993), the expected value or mean number of the people who carry the defective gene in the sample of 1000 individuals would be equal to 5.
Hence, the approximate distribution of the number who carry this gene corresponding to this sample is Poisson distribution: Poisson(5). That is, the pmf of the rv $X$ is
\[
p(x;\lambda)=p(x;5)=\frac{e^{-5}5^x}{x!},\quad x=0,1,2,\ldots
\]
(a) The approximate probability that $X$ between $5$ and $8$ (inclusive) carry the defective gene is
\[
P(5\leqslant X\leqslant 8)=\sum_{x=5}^{8}p(x;5)=\sum_{x=5}^{8}\frac{e^{-5}5^x}{x!}\approx 0.491413
\]
Do calculation by Calculator.exe
>> sum(5^x/(x!),x,5,8)
in> sum(5^x/(x!),x,5,8)
out> 72.932168
------------------------
>> exp(5)
in> exp(5)
out> 148.413158
------------------------
>> 72.932168/148.413158
in> 72.932168/148.413158
out> 0.491413
(b)
The approximate probability that $X$ will be at least $8$ (inclusive) who carry the defective gene is
\[
\begin{split}
P(X\geqslant 8)&=1-P(X < 8)=1-P(X\leqslant 7)\\
&=1-\sum_{x=0}^{7}p(x;5)=1-\sum_{x=0}^{7}\frac{e^{-5}5^x}{x!}\\
&\approx 1-0.866628\\
&=0.133372
\end{split}
\]