5. In a Cartesian coordinate system, points with both coordinates being integers are called integer points, and the set of all integer points is denoted as $X$. Find the largest real number $\alpha$, such that there exists a function $f: X \rightarrow \mathbf{N}^{*}$, satisfying: (1) There exists a positive real number $c$, such that for each positive integer $n$, the number of integer points $A$ satisfying $f(A)=n$ is greater than $c n^{\alpha}$; (2) For any integer point $\triangle A B C$ (i.e., $A, B, C$ are all integer points), $f(A)>|B C|, f(B)>|A C|$, $f(C)>|A B|$ at least one of these conditions holds.
【Answer】The maximum value sought is $\alpha=1$. (3 points)
First, we show that such an $f$ exists when $\alpha=1$. Let $c=\frac{1}{2}$, and for point $A=(a, b)$, we define $f(a, b)=2|a|+2|b|+t$, where $t \in\{0,1\}$ and $t+1 \equiv a(\bmod 2)$.

For such an $f$, it is clear that $f(A) \geq 2|O A|$. (6 points)
For condition (1): $f(O)=1$, on the even sequence $2 k$ there is one odd number $|k|+1$ and two larger odd numbers, while on the odd sequence, they are all even numbers. It is easy to see that the number of points satisfying $f\left(A_{i}\right)=n$ is at least $\frac{n}{2}$. (9 points) For condition (2):
Assume there is an integer point triangle $A B C$ that does not meet the requirements. Without loss of generality, assume $A B$ is the longest side, then
$$
f(A)+f(B) \geq 2|O A|+2|O B| \geq 2|A B| \geq|B C|+|A C| \text {. }
$$

In this case, all equalities must hold, indicating that $\triangle A B C$ is an equilateral triangle, and $O$ is simultaneously on $A B, B C, C A$, which is impossible. (12 points) Therefore, when $\alpha=1$, such an $f$ meets the requirements.
【Explanation】Next, we prove that $\alpha>1$ does not meet the requirements. Assume there exists an $f$ that meets the requirements at this time.
For any point $X$, satisfying $|X A|<n$, for any other point $X$ in $M$, it either lies on the line $A B$, or one of $|X A|,|X B|$ is no more than $n$, thus within the circle with $A, B$ as centers and radius $n$. There are no more than $8 n^{2}$ points within the circle. (15 points)
If there are more than $12 n$ points on the line $A B$, according to the pigeonhole principle, there must exist three points with pairwise distances all no less than $4 n$. At this time, any point outside the line will have a distance greater than $n$ to at least two of these points, contradicting the fact that all points in $M$ are not collinear! Therefore, $|M|\sum_{k=1}^{n} c \cdot k^{\alpha} \geq \sum_{k=\frac{n}{2}}^{n} c \cdot k^{\alpha} \geq c \cdot\left(\frac{n}{2}\right)^{\alpha+1}$, if $\alpha>1$, when $n$ is sufficiently large, $c \cdot\left(\frac{n}{2}\right)^{\alpha+1}-1>8 n^{2}+12 n$, leading to a contradiction!
$(21$ points)
In summary, $\alpha=1$ is the maximum value sought.