$\bullet$ In $\mathbb Z[\sqrt{-5}]$ why is $(2)=(2,1+\sqrt{-5})(2,1-\sqrt{-5})$
Actually both ideals on the RHS contain $(2)$, but also their product ?
Can we just multiply RHS in the normal sense;
$2\cdot2=4,\quad \left(1+\sqrt{-5}\right)\left(1-\sqrt{-5}\right)=6$
Hence $(2)=(6-4)\subset(4,6)\subset(2,1+\sqrt{-5})(2,1-\sqrt{-5})$
It remains to show the reverse containment...
$\bullet$ Now the second question: What is the norm of $(2,1+\sqrt{-5})$ or $(2,1+\sqrt{-5})$, using the multiplicativity of the norm one of them must have $2$, and the other $1$ ? (since $(2)$ has norm $2$)
You have that $(2, 1+\sqrt{-5})(2, 1-\sqrt{-5})$ is the ideal generated by products of generators, i.e. it is $(4, 2+2\sqrt{-5}, 2-2\sqrt{-5}, 6)$. Since all of its generators belong to $(2)$, you have the other inclusion.
As for the second question, recall that $N((2))=N(2)=2^2=4$. The two ideals (call them $I,J$) must satisfy $N(I)N(J)=N(IJ)=N((2))=4$. Actually, you have $I=J$ because $$1-\sqrt{-5}=2-(1+\sqrt{-5})$$ $$1+\sqrt{-5}=2-(1-\sqrt{-5})$$ so you have $N(I)^2=4$, hence $N(I)=2$.