Part 1 is still kind of nice; depending on what you need this for this may or may not be useful:
For a real polynomial, its roots are either real or form conjugate pairs. If you have two conjugate roots of unity
$$e^{\frac{2i\pi k}{n}},\ e^{-\frac{2i\pi k}{n}},$$
then the quadratic polynomial with them as roots is
$$P(x)=x^2-2\cos\left(\frac{2\pi k}{n}\right)x+1.$$
So, any real-coefficiented polynomial (whose roots are all roots of unity) has the form
$$(x+1)^m(x-1)^n\prod_{k=1}^N \left(x^2-2\cos\left(\frac{2\pi a_k}{b_k}\right)x+1\right),$$
where $m,n$ are nonnegative integers and all $a_k,b_k$ are positive integers (with $a_k<b_k$, if you wish).
For part 2, I mean, you can just pick the roots to be
$$x_k=e^{\frac{2i\pi a_k}{b_k}},$$
and your polynomial will be
$$\prod_{k=1}^N \left(x-e^{\frac{2i\pi a_k}{b_k}}\right),$$
but this probably isn't that useful. Another thing that may or may not be useful for either of these is that all polynomials with only roots of unity as roots have roots that all satisfy $x^M=1$ for some (possibly large with respect to the degree of the polynomial) integer $M$. In the case above this $M$ can be taken to be $b_1b_2\cdots b_k$. So, all polynomials with only roots of unity as roots are factors of $x^M-1$ for some $M$.
I'll put here what I left as a comment above.
If $k$ is the residue field of $K$ of characteristic $p$, then there is an isomorphism (of groups) $k^{\times} \rightarrow \mu_K^{(p)}$, where $\mu_K^{(p)}$ is the set of roots of unity in $K$ whose order is not divisible by $p$. (This isomorphism is given by the Teichmuller map).
So all that remains is to check the roots of unity of two power order (in fact it suffices to show $i=\sqrt{-1} \not\in K$ which you can do via ramification theory).
Best Answer
Note that first of all $\mathcal O_{\mathbb Q(\sqrt{-5})}^\times=\mathbb Z[\sqrt{-5}]^\times$ is very small. If $a+b\sqrt{-5}\in\mathbb Z[\sqrt{-5}]^\times$ then $N_{\mathbb Q(\sqrt{-5})/\mathbb Q}(a+b\sqrt{-5})=a^2+5b^2=\pm1$, which only happens when $a=\pm1$ and $b=0$. Thus $\mathcal O_{\mathbb Q(\sqrt{-5})}^\times=\{\pm1\}$.
P.S. The argument generalizes to $\mathbb Q(\sqrt{-d})$ for $d>4$. When $d\equiv1,2\pmod 4$ the same exact argument applies. When $d\equiv3\pmod 4$, you have $N_{\mathbb Q(\sqrt{-d})}(a+b\frac{1+\sqrt{-d}}2)=a^2+ab+\frac{1+d}4b^2=\pm1$, which you can show only happens when $a=\pm1$ and $b=0$.