Answer:
a) There is a 12.11% probability that exactly 1 man has the marker.
b) There is a 85.07% probability that more than 1 has the marker.
Step-by-step explanation:
There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.
Binomial probability
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which
is the number of different combinatios of x objects from a set of n elements, given by the following formula.

And
is the probability of X happening.
In this problem, we have that:
30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so 
(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?
10 men, so 
We want to find P(X = 1). So:


There is a 12.11% probability that exactly 1 man has the marker.
(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?
That is 
We have that:


We also have that:


So

Finally

There is a 85.07% probability that more than 1 has the marker.