1) By my course, $\mathbb Z[t]$ is a unique factorisation domain. But I don't understand since $$10=2\cdot 5=(-2)\cdot (-5)=(-1)\cdot (2)\cdot (-5)=(-1)\cdot (-2)\cdot (-5)$$ which are different factorization, isn't it ? So how could we precise that we don't care about the sign ? The only thing I have is that if $$wp_1\cdot ...\cdot p_n=v q_1\cdot ...\cdot q_r$$ then $r=s$ and that there exist a permutation such that $p_i=q_{\sigma (t)}$ for all $i$. So, how to manage with the sign ?
2) To show that $\mathbb Z[\sqrt{-5}]$ is unique domain factorization, I have that $$6=(1+\sqrt{-5})\cdot (1-\sqrt{-5})=2\cdot 3$$ but how can I show that $1+\sqrt{-5}$, $1-\sqrt{-5}$, $2$ and $3$ are irreducible ? I would like to say that $$\mathbb Z[\sqrt{-5}]=\{a+b\sqrt{-5}\mid a,b\in\mathbb Z\},$$
and since $2$ and $3$ are irreducible in $\mathbb Z$ it is also in $\mathbb Z[\sqrt{-5}]$ (but I'm not sure if this argument is true). For $1+\sqrt{-5}$ and $1+\sqrt{-5}$ I don't know how to justify.
1) When defining UFD, we does not distinguish multiple by units. That is, $p_1p_2\cdots p_n$ and $up_1p_2\cdots p_n$ are same factorization when $u$ is a unit. In your example, you can get $2\cdot 5$ from $(-1)(-2)(-5)$ by multiplying $(-1)^3$.
2) Your argument for proving irreducibility of $2$ is not valid. For example, 2 is irreducible in $\Bbb{Z}$, but it is not irreducible in $\Bbb{Z}[i]$, since $2=(1+i)(1-i)$. However you can prove it in other way, by applying the definition of irreducible elements directly. I will give a proof of irreducibility of $2$. Other cases ($3$ and $1\pm \sqrt{-5}$) are same.
If $2$ is irreducible, we can find integers $a$, $b$, $c$, $d$ s.t. $2=(a+b\sqrt{-5})(c+d\sqrt{-5})$. Such $a$, $b$, $c$, $d$ should satisfy $$ac-5bd = 2 \quad\text{and}\quad ad+bc=0$$ It forces $bd=0$, so $ac=2$. Without loss of generality we can assume that $b=0$. Then $ad=0$ and $ac=2$. If $a=0$ then $ac=0$, so $a\neq 0$ and $d=0$. From $ac=2$, we get one of $a$ and $c$ must be a unit, so one of $a+b\sqrt{-5}$ and $c+d\sqrt{-5}$ is a unit.