You can consider how primes split in $\mathcal{O}_K=\mathbb{Z}(\sqrt{-5})$. If we try to find the discriminant since $-5 \equiv 3\pmod 4$, $D= 4 \cdot -5=-20$, and primes not dividing $-20$ split if $\left(\frac{-20}{p}\right)=\left(\frac{-4}{p}\right)\left(\frac{p}{5}\right)=1$. From there you should be able to get a set of congruence conditions for primes that split. As far as primes that don't split, you only need consider those primes $<10$, since otherwise $N(p)=p^2>100$. Finally $2$ and $5$ both ramify so primes of $\mathcal{O}_K$ lying above them also need to be considered but should be a short, finite check from there.
I get $\{2, 3, 5, 7, 23, 29, 37, 41, 43, 47, 61, 67, 83, 89\}$ as the list of primes with norm less than $100$ that ramify or split using the congruences $p \equiv 1, 3, 7, 9 \pmod {20}$ for primes that split in $\mathcal{O}_K$. There are no inert primes with norm $<100$.
Here is a way to proceed without too much algebraic number theory. Of course, I assume you know that $O_K=\mathbb{Z}[\sqrt{10}]$ (see all the comments)
Of course, the unlerlying idea is that $I$ has norm $2$, and that there is no element in $O_K$ having norm $\pm 2$, but we don't need that.
In what follows, $N=N_{K/\mathbb{Q}}$.
Assume that $I=(\alpha)$ for some $\alpha\in O_K=\mathbb{Z}[\sqrt{10}]$. So $\alpha\mid 2$ and $\alpha\mid \sqrt{10}$ in $O_K$. Since norms are multiplicative, taking the norms yields $N(\alpha)\mid 4$ and $N(\alpha)\mid -10$ in $\mathbb{Z}$, so $N(\alpha)\mid 2$. It follows that $N(\alpha)=\pm 1,\pm 2$. The case $N(\alpha)=\pm 1$ implies that $\alpha$ is a unit, and that $I=O_K$. I let you check this is not the case.
So $N(\alpha)=\pm 2$. Writing $\alpha=a+b\sqrt{10}$, we have to solve $a^2-10 b^2=\pm 2$. In particular, $\pm 2$ is a square in $\mathbb{Z}/10\mathbb{Z}$. You can check this is not the case. Hence $I$ is not principal.
Side comment: if you want to show that $N(I)=2$, then you have several ways to proceed, depending what you know in algebraic number theory.
First method (very elaborate): since $2\in I$, you can try to factor $(2)$ as a product of prime ideals and see what happens. Since $O_K=\mathbb{Z}[\sqrt{10}]$, the factorisation a prime number $p$ is reflected by the factorisation of $X^2-10$ mod $p$. Here, $X^2-10=X^2$ mod $p$, and then $(2)=(2,\sqrt{10})^2=I^2$ by a famous theorem of Dedekind.
In particular, $N(I^2)=N(I)^2= N(2O_K)=\vert N(2)\vert =4$, so $N(I)=2$.
Second method(elementary): Well, you can check that $I=\{a+b\sqrt{10}, a\in 2\mathbb{Z},b\in\mathbb{Z}\}$. This is not totally obvious, and does not comes directly from the definition of $I$, since a generic element of $I$ has the form $2z_1+\sqrt{10}z_2,$ with $z_1,z_2\in O_K $(and not $z_1,z_2\in\mathbb{Z}$), so this requires a bit of easy computation. Hence, a $\mathbb{Z}$-basis of $O_K$ is $(1,\sqrt{10})$, while a basis of $I$ is $(2,\sqrt{10})$. Thus, as an abelian group, we get $O_K/I\simeq \mathbb{Z}/2\mathbb{Z}$ and $N(I)=\vert O_K/I\vert =2$.
Best Answer
Re your first question....
By choosing a $\mathbb{Z}$-basis $\{ 1, 1 + \sqrt{-5} \}$ for $\mathcal{O}_K$, we can write algebraic integers as coordinate "vectors".
The generators of the ideal $I$ are, relative to this basis, $(2,0)$ and $(0, 1)$. This turns out to be a $\mathbb{Z}$-basis for $I$, although you might want to double check by throwing in all four elements of $\{ 2, 1 + \sqrt{-5} \} \cdot \{ 1, \sqrt{-5} \}$ into a matrix and doing (integer) row reduction to simplify. (using all four of these elements ensures that their span is closed under multiplication by elements of $\mathcal{O}_K$).
One definition of the norm of an ideal is that it is the size of the quotient ring $\mathcal{O}_K / I$. Knowing that the basis for $I$ (relative to the basis for $\mathcal{O}_K$) has coordinate matrix
$$ \left( \begin{matrix}2 & 0 \\ 0 & 1 \end{matrix} \right)$$
makes it easy to see that the quotient group has two elements.
Actually, when we do things this way, we can obtain the norm of $I$ as the determinant of its basis matrix.