I need to know if my approach to the problem in the title is correct or not.
First, what is $(n,k,d)-code$? It is a (binary) code $C$, $|C|=2^k$, $\Delta(C)=d$ (that is Hamming's distance).
So what is the minimum $n$ that an $(n,4,3)-code$ exists? First I tried using Singleton's bound but I am not quite sure what does it tell me and when. I am kind of confused about it. After that I tried using Hamming's bound with $n=6$. That gives us $$2^4 \leq \frac{2^6}{\binom{6}{0}+\binom{6}{1}}$$ which is false so I assume that no $(6,4,3)$ code exists. For $n=7$ it is not false so such code could exist but still I cannot be sure. That is where I use Gilbert-Varsham's bound. So for $n=7$ it gives us: $$2^4 \geq \frac{2^7}{\binom{7}{0}+\binom{7}{1}+\binom{7}{2}}$$ which is true so I can assume that the minimum n for a valid binary $(n,4,3)-code$ is $7$.
Is this correct and can I apply the same logic and same proving for the $(n,3,3)-code$?
The bounds you cite give a lower limit for $n$. They do not guarantee that $n$ will work. One way to prove that a given $n$ will work is to display such a code. The $(7,4,3)$ code is available for lookup as it is a common example. You can then generate a $(6,3,3)$ code from it by removing all the words that have a $1$ in one of the data bits and deleting that data bit.