Randomly selecting a natural number
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)
$$\lim_{N} \frac{\text{amount of even natural numbers} \leq N}{N} = P( 0 \mod 2) =\frac{1}{2} \,.$$
Perhaps a justification is this. In the first question it is (correctly) claimed that it is impossible to have a uniform distribution on the natural numbers. Thus we can't develop a sensible way of choosing particular numbers at random, when each supposedly has the same probability. The last question though is dealing with the probability of picking a certain class of numbers. In that case the approach is to pick $N$ large, impose a uniform distribution on $[1,N]$ (which we can always do), work out the probability as a function of $N$, and then take the limit $N \rightarrow \infty.$
Example: what's the probability of randomly picking an even number?
Fix $N$. If $N$ is even then the probability of picking an even number in the uniform distribution on $[1,N]$ is exactly $1/2$. If $N$ is odd then the probability is $$ \frac{\text{amount of even numbers}}{\text{total amount}} = \frac{ (N-1)/2 }{N} = \frac{1}{2} - \frac{1}{2N} $$ And so, as we take the limit $N \rightarrow \infty$, we say that the probability of choosing an even number is 1/2.
This approach is really a "natural density" approach; see http://en.wikipedia.org/wiki/Natural_density.
The natural density features in, for instance, the Green-Tao theorem. Green and Tao showed that the primes have positive natural density, and thus that they must contain arbitrarily long arithmetic sequences. <- incorrect!!