Why this happen only with $4,8,12,.......$ We will take some examples to illustrate my question:
If we take a set of numbers,  for example $1,2$, and $3$
$$1+2^4+3^4=98$$
$$1+2^8+3^4=338$$
$$1+2^8+3^8=6818$$
$$1+2^8+3^{12}=531698$$
We note that all numbers resulting end 8.This will not change for any selected power from the specified values(4,8,12....) in question.
another example
$$2^4+4^4+7^4+10^4=12673$$
$$2^4+4^8+7^8+10^4=5840353$$
$$2^{16}+4^8+7^8+10^4=5905873$$
$$2^{16}+4^8+7^8+10^8=105895873$$
We note that all numbers resulting end $3$
My question "Why this happen only with these numbers and how can I prove this?"

Because for any integer $x$ whatsoever, and any $r,s\geq 1$,
$$x^{4r}\equiv x^{4s}\bmod 10$$
As an example, with $r=3$, and $s=5$ this says that
$$1^{12}\equiv 1^{20}\bmod 10,\qquad 2^{12}\equiv 2^{20}\bmod 10,\qquad 3^{12}\equiv 3^{20}\bmod 10$$
and therefore
$$\begin{align*}
(\text{last digit of $1^{12}+2^{12}+3^{12}$})&\equiv1^{12}+2^{12}+3^{12}\\
&\equiv 1^{20}+2^{20}+3^{20} \equiv (\text{last digit of $1^{20}+2^{20}+3^{20}$})\bmod 10
\end{align*} $$
To see that my claim is true, note that
$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|}
x   & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\\hline
x^4\bmod 10 & 0 & 1 & 6 & 1 & 6 & 5 & 6 & 1 & 6 & 1
\end{array}$$
and modulo $10$, the numbers $0,1,5,6$ are all fixed by $x\mapsto x^k$ for any $k$.
