It really depends on what you mean by the "probability of randomly selecting n natural numbers with property $P$". While you cannot pick random natural number, you can speak of uniform distribution.
For the last problem, the probability is calculated, and is to be understood as the limit when $N \to \infty$ from the "probability of randomly selecting n natural numbers from $1$ to $N$, all pairwise coprime".
Note that in this sense, the second problem also has an answer. And some of this type of probabilities can be connected via dynamical systems to an ergodic measure and an ergodic theorem.
Added The example provided by James Fennell is good to understand the last paragraph above.
Consider ${\mathbb Z}_2 = {\mathbb Z}/2{\mathbb Z}$, and the action of ${\mathbb Z}$ on ${\mathbb Z}_2$ defined by
$$m+ ( n \mod 2)=(n+m) \mod 2$$
Then, there exists an unique ergodic measure on ${\mathbb Z}_2$, namely $P(0 \mod 2)= P(1 \mod 2)= \frac{1}{2}$.
This is really what we intuitively understand by "half of the integers are even".
Now, the ergodic theory yields (and is something which can be easily proven directly in this case)
$$\lim_{N} \frac{\text{amount of even natural numbers} \leq N}{N} = P( 0 \mod 2) =\frac{1}{2} \,.$$
This is an interesting question. I don't have a real answer but I wrote a program to calculate values for different $n$ and $m$, counting the number of ways to split the first $n$ numbers into groups of arithmetic sequences of size $m$ . Here are some results with the pairs are written as $(n,m)$:
$(9,3)=5$,
$(12,3)=15$, $(12,4)=4$,
$(15,3)=55$, $(15,5)=4$,
$(16,4)=11$,
$(18,3)=232$, $(18,6)=4$, $(18,6)=4$,
$(20,4)=23$, $(20,5)=10$,
$(24,3)=6643$, $(24,4)=68$, $(24,6)=10$, $(24,8)=4$.
I skipped writing pairs that satisfy these equations:
If $m$ is not a divisor of $n$ then $(n,m)=0$. If $n$ is even then
$$(n,\frac{n}{2})=2,$$
$$(n,2)=\frac{n!}{2^{n/2}\frac{n}{2}!}.$$
EDIT
This wouldn't fit in a comment, but here are the ways of grouping (12,3):
$$
\begin{array}{ccc}
\left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \\
10 & 11 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 9 & 11 \\
8 & 10 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 6 & 8 \\
5 & 7 & 9 \\
10 & 11 & 12
\end{array}\right]\\
\left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 7 & 10 \\
5 & 8 & 11 \\
6 & 9 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 8 & 12 \\
5 & 6 & 7 \\
9 & 10 & 11
\end{array}\right] & \left[\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
7 & 8 & 9 \\
10 & 11 & 12
\end{array}\right]\\
\left[\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
7 & 9 & 11 \\
8 & 10 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 3 & 5 \\
2 & 6 & 10 \\
4 & 8 & 12 \\
7 & 9 & 11
\end{array}\right] & \left[\begin{array}{ccc}
1 & 3 & 5 \\
2 & 7 & 12 \\
4 & 6 & 8 \\
9 & 10 & 11
\end{array}\right]\\
\left[\begin{array}{ccc}
1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9 \\
10 & 11 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 5 & 9 \\
2 & 3 & 4 \\
6 & 7 & 8 \\
10 & 11 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 5 & 9 \\
2 & 4 & 6 \\
3 & 7 & 11 \\
8 & 10 & 12
\end{array}\right]\\
\left[\begin{array}{ccc}
1 & 5 & 9 \\
2 & 6 & 10 \\
3 & 7 & 11 \\
4 & 8 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 6 & 11 \\
2 & 3 & 4 \\
5 & 7 & 9 \\
8 & 10 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 6 & 11 \\
2 & 7 & 12 \\
3 & 4 & 5 \\
8 & 9 & 10
\end{array}\right]
\end{array}
$$
Best Answer
"From first $100$ natural numbers, $3$ numbers are selected. If these three numbers are in A.P., then find the probability that these numbers are even."
According to this text a conditional probability is asked for. Therefore we do not have to count the possible triple selections from $[100]$, but only the possible A.P. triple selections from $[100]$, and to analyze how many of these are all-even triples.
We obtain an A.P. triple by choosing either two even numbers and their average or two odd numbers and their average. Therefore we obtain $2\cdot{50\choose2}=2450$ possible A.P. triples, all of them equally probable.
An all-even A.P. triple can be halved termwise, and in this way becomes an A.P. triple from $[50]$. There are $2\cdot{25\choose2}=600$ such triples.
The probability $p$ you are after therefore is given by $$p={600\over2450}={12\over49}\ .$$