In the answer for What is p-adic logarithmic map of an elliptic curve? How to compute it? @alex-j-best says:
sage: pP,pQ # points near infinity we can take log of
pP = $((4a + 4)5^-2 + (a + 1) + (2a + 2)5 + (a + 1)5^2 + (2a + 3)5^3 + (2a + 2)5^4 + (3a + 3)5^5 + (4a + 1)5^6 + (a + 2)5^7 + (a + 1)5^8 + 4a5^9 + (3a + 4)5^10 + 3a5^11 + (a + 4)5^12 + a5^13 + (4a + 3)5^14 + (a + 2)5^15 + O(5^17) : (4a + 3)5^-3 + (4a + 2)5^-2 + (a + 2)5^-1 + (4a + 4) + 2a5 + (a + 1)5^2 + 5^3 + 5^4 + (2a + 2)5^5 + (4a + 1)5^6 + (a + 3)5^7 + 35^8 + (4a + 1)5^9 + 25^10 + (2a + 2)5^11 + (a + 1)5^12 + (3a + 3)5^13 + 2a5^14 + O(5^16) : 1 + O(5^20))$
pQ = $((a + 1)5^-2 + (4a + 4)5^-1 + (4a + 4) + a5 + (4a + 1)5^2 + (2a + 3)5^4 + (2a + 3)5^5 + (3a + 2)5^6 + (3a + 3)5^7 + (3a + 4)5^8 + 3a5^9 + (4a + 3)5^10 + (3a + 1)5^11 + (a + 4)5^12 + (3a + 4)5^13 + (3a + 3)5^14 + (4a + 2)5^15 + (a + 1)5^16 + O(5^17) : (3a + 1)5^-3 + (3a + 3)5^-2 + (3a + 2)5^-1 + (2a + 1) + 45 + (4a + 3)5^2 + (3a + 2)5^3 + (2a + 1)5^4 + (4a + 3)5^5 + (4a + 4)5^6 + (a + 3)5^7 + (3a + 3)5^9 + 35^11 + 2a5^13 + a5^14 + (4a + 2)5^15 + O(5^16) : 1 + O(5^20)))$
why are those points near to point at infinity?
The short answer to your question is that a point is near infinity if its ordinary $(x,y)$-coordinates are large. In the case of your specific points, here’s what I say:
It looks as if you’re using the $5$-adic metric, and it looks (granting that your post still needs editing to make its MathJax grammatical) as if your first point is $\bigl(\frac{4a+4}{5^2}+\cdots:\frac{4a+3}{5^3+\cdots}:\,1 \bigr)$. That is, the point is given in homogeneous coordinates $(x:y:z)$.
Multiplying through by $5^3$, we get an equivalent description of your point as $\bigl(5(4a+4)+\cdots:(4a+3)+\cdots:5^3\bigr)\,.$ Now, the “point at infinity” has coordinates $(0:1:0)$, so you see, thinking $5$-adically, that your (first) point is close to infinity. Same story for the second point.
Sage did you dirt by giving you far too much information.
EDIT: Addition
You have asked why $\bigl(5(4a+4)+\cdots:(4a+3)+\cdots:5^3\bigr)$ is like $(0:1:0)$. These are homogeneous coordinates, so if we call $u$ the middle coordinate I’ve just mentioned, then you have a homogeneous description of our point as $$ \left(5\frac{4a+4+\cdots}u:1:\frac{5^3}u \right)\,, $$ after dividing all three coordinates by $u$. Then the first coordinate is divisible by $5$ and the third is exactly divisible by $5^3$, since $u$ is a $5$-adic unit. Divisible by powers of $5$ means small, i.e. close to zero.