(I): in $\mathbf Z[\sqrt{-5}]$ , the classic factorisation $6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5})$ shows $3\mid(1+\sqrt{-5}) (1-\sqrt{-5})$ . Using the norm $N(a+b\sqrt{-5})=a^2+5b^2$ : $N(3)=9$ and $N(1\pm\sqrt{-5})=6$ ; if $3$ divided either factor, $9=N(3)$ would have to divide $N(1\pm\sqrt{-5})=6$ , which it does not. So $3$ divides the product but neither factor — $3$ fails the defining property of a prime element. (It is still irreducible: $a^2+5b^2=3$ has no integer solutions, so $3$ cannot split into two non-unit factors.) Statement (I) is false.

(II): in $\mathbb{Z}/10\mathbb{Z}$ , the ideal $(\bar2)$ has quotient $(\mathbb{Z}/10\mathbb{Z})/(\bar2)\cong\mathbb{Z}/2\mathbb{Z}$ , an integral domain, so $(\bar2)$ is a prime ideal and $\bar2$ is a prime element. For irreducibility, look for a factorisation into two non-units: units mod $10$ are $\{1,3,7,9\}$ , and $6\cdot2=12\equiv2\pmod{10}$ expresses $\bar2$ as a product of the two non-units $\bar6$ and $\bar2$ — so $\bar2$ is not irreducible. Statement (II) is true.