Question: Test of divisible by $4 ,$ last two digit must be divisible by $4 ,$ since $100$ is always divisible by $4$ remaining two digit $,$ we need to check $.$

Test of divisible by $3 ,$ sum digits must be divisible by $3 .$ But I don't know why $?$

Test of divisible by $2 ,$ last digit must be divisible by $2 ,$ since $10$ is always divisible by $2 .$

Test of divisible by $ 6 ,$ the number should be divisible by $2$ and $3 .$

Test of divisible by $12 ,$ the number should be divisible by $4$ and $3 ,$ somewhere is given as we can say if number is divisible by $2$ and $6$ then it also must be divisible by $12 ,$ for counter example $,$ if number is $18 .$

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I have two questions $:$

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> 1. Why test of divisible by $3$ works $?$ Any proof $.$
> 2. Why test of divisible by $12$ works with divisible by $3$ and $4 ,$ but not with divisible by $2$ and $6 ?$ Any proof $.$
> 
> 
>

Answer: 
The reason the divisibility by $3$ test works is that $10$ and $1$ have the same remainder when divided by $3$. Say your number is (as digits) $x = a\_n a\_{n-1} \ldots a\_1 a\_0$. Then you could write it as $x = a\_n \cdot 10^n + a\_{n-1} 10^{n-1} + \cdots + a\_1 10 + a\_0$. If you mod out by $3$:
$$
\begin{align\*}
x \bmod 3 &= (a\_n \cdot 10^n + a\_{n-1} 10^{n-1} + \cdots + a\_1 10 + a\_0) \bmod 3 \\
&= (a\_n \cdot 10^n) \bmod 3 + (a\_{n-1} 10^{n-1}) \bmod 3 + \cdots + (a\_1 10) \bmod 3 + a\_0 \bmod 3) \\
&= (a\_n \bmod 3) \cdot (10^n \bmod 3) + (a\_{n-1} \bmod 3) (10^{n-1} \bmod 3) + \cdots + (a\_1 \bmod 3) \cdot (10 \bmod 3) + (a\_0 \bmod 3) \\
&= (a\_n \bmod 3) \cdot 1 + (a\_{n-1} \bmod 3) \cdot 1 + \cdots + (a\_1 \bmod 3) \cdot 1 + (a\_0 \bmod 3) \\
&= (a\_n \bmod 3) + (a\_{n-1} \bmod 3) + \cdots + (a\_1 \bmod 3) + (a\_0 \bmod 3) \\
&= (a\_n + a\_{n-1} + \cdots + a\_1 + a\_0) \bmod 3
\end{align\*}
$$

So $x$ and the sum of its digits have the same remainder when divided by $3$.

As for your second question: if you know $x$ is divisible by $6$, you already know it's divisible by $2$. Doing the test for $2$ doesn't give you any new information. This is because $6$ and $2$ aren't relatively prime (but $3$ and $4$ are). In general, if you know $x$ is divisible by $a$ and $b$, it's divisible by the LCM of $a$ and $b$.
Since we are working in a numerical system of basis $10$, an integer number $N>0$ can be written as $$N=a\_n\times 10^n+a\_{n-1}\times 10^{n-1}+\ldots +a\_1\times 10 +a\_0$$
The number $10=9+1$ leaves remainder $1$ under division by $3$, and so does the another powers of $10$, so $N$ has the form $$N=9m+a\_{n}+a\_{n-1}+\ldots+a\_1+a\_0$$
for some integer $m$, which means that $3$ divides $N$ iff $3$ divides $a\_{n}+a\_{n-1}+\ldots+a\_1+a\_0$, the sum of the digits of $N$. Also the same test works for $9$.