thanks for taking the time to look at my problems.

I was trying to calculate the norm of $(3, 1 + \sqrt{-17})$ and $(\sqrt{-17})$.
The second one is 17 because of the norm of the element $\sqrt{-17}$, but how does this follow from $|\mathbb{Z}[\sqrt{-17}]/(\sqrt{-17})|$?

I tried to calculate $|\mathbb{Z}[\sqrt{-17}]/(3, 1+\sqrt{-17})|$ and concluded that $\mathbb{Z}[\sqrt{-17}]/(3, 1+\sqrt{-17}) \cong \mathbb{Z}/3\mathbb{Z}$ such that $|\mathbb{Z}[\sqrt{-17}]/(3, 1+\sqrt{-17})| = 3$. Is this correct?

Thanks in advance!
['https://math.stackexchange.com/questions/1803867', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/296749/']
Your computations are correct.

Since $-17\equiv3\bmod{4}$ our ring of integers is $\mathbb{Z}[\sqrt{-17}]$, so we may factor the ideal $(3)$ in $\mathbb{Z}[\sqrt{-17}]$ by factoring 

$$x^2 + 17 \equiv x^2 - 1 \equiv (x+1)(x+2) \bmod{3}.$$ 

This yields the ideal $(3,1+\sqrt{-17})$, and since 3 splits the norm of this ideal is 3. 

To see this more directly, we can use the ring isomorphism theorems. We have $(x^2+17) \subseteq (3,1+x) \subseteq \mathbb{Z}[x]$ from above, hence 

$$\mathbb{Z}[\sqrt{-17}]/(3,1+\sqrt{-17}) \cong
 \mathbb{Z}[x]/(3,1+x) \cong
 \mathbb{Z}/3$$

As for $(\sqrt{-17})$, the same argument works.
You already seem know that the norm of a prinicipal ideal is the norm of its generator. Hence $|\mathbb Z[\sqrt{-17}]/(3)|=9$.

We have $(3) \subsetneq (3,1+\sqrt{-17}) \subsetneq (1)$, hence $\mathbb Z[\sqrt{-17}]/(3,1+\sqrt{-17})$ is a non-trivial quotient of $\mathbb Z[\sqrt{-17}]/(3)$. A non-trivial quotient of a group with $9$ elements must have $3$ elements, hence we obtain

$$|\mathbb Z[\sqrt{-17}]/(3,1+\sqrt{-17})|=3$$

without having calculated the quotient (Which would be of course an easy task, too, as shown in the other answers).