Is the language $a^{3}b^{+}$ the same as $\{a^{3}b^{n}, n \geq 1\}$ ? and what is the result of pumping this?
You are making a classical mistake in applying the pumping lemma. The pumping lemma states that each word $w$ of length at least $N$ in the language (where $N$ is the number of states in the minimal DFA for the language) can be decomposed as $w = xyz$, where $|xy| \leq N$, $|y| \geq 1$, and $xy^i z$ is in the language for all $i$. You don't get to choose this partition! The lemma just states that there is some partition. Taking as an example your word $a^3 b^t$, since $N = 6$ (indeed, the minimal DFA for the language has $6$ states), we can always choose $x = a^3$ and $y = b$, and then $xy^i z = a^3 b^{t+i-1}$ which is always in the language (since $t \geq 3$).

When you use the pumping lemma to prove that a language is not regular, you don't know $N$, and so you have to start by taking $N$ to be arbitrary, and run the proof from there. What you are in effect showing is that for each $N$, there is no DFA accepting the language with at most $N$ states.