4. First, we prove a lemma:
Lemma 2. $\prod_{k=0}^{m n-1}\left(1+e^{2 \pi i k / n}\right)=\left(1-(-1)^{n}\right)^{m}$.
Proof of the lemma: When $m=1$, since $1-x^{n}=\prod_{i=0}^{n}\left(e^{2 \pi i k / n}-x\right)$, substituting $x=-1$ proves the lemma. If $m>1$, since $e^{2 \pi i k / n}=e^{2 \pi i(k-m) / n}$, we actually have:
$$
\prod_{k=0}^{n m-1}\left(1+e^{2 \pi i k / n}\right)=\prod_{l=1}^{m} \prod_{k=(l-1) n}^{m-1}\left(1+e^{2 \pi i k / n}\right)=\prod_{l=1}^{m} \prod_{k=0}^{n-1}\left(1+e^{2 \pi i k / n}\right)=\left(1-(-1)^{n}\right)^{m}
$$

Now, we consider $\prod_{b=0}^{2018}\left(1+e^{\frac{2 \pi a n k t}{2014}}\right)$. If $a$ is coprime with 2019, then $0 a, a, 2 a, \ldots, 2018 a$ is also a complete residue system modulo 2019, so by the lemma we have:
$$
\prod_{b=0}^{2018}\left(1+e^{\frac{2 \pi a b k}{2019}}\right)=\prod_{b=0}^{2018}\left(1+e^{\frac{2 \pi 1 b}{2019}}\right)=2^{(a, 2019)}
$$

If $a$ is not coprime with 2019, since all factors of 2019 are odd, by the lemma we get:

So

Then
$$
\log _{2}\left(\prod_{a=0}^{2018} \prod_{b=0}^{2018}\left(1+e^{\frac{2 \pi n a b}{2019}}\right)\right)=2019+\sum_{a=1}^{2018}(a, 2019)
$$

Since $2019=3 \times 673$, we have $\sum_{a=1}^{2018}(a, 2019)=3 \times(2019 \div 3-1)+673 \times(2019 \div 673-1) + 1 \times(2018-(2019 \div 3-1)-(2019 \div 673-1))=3 \times 672+673 \times 2+1344=4706$, so the original expression $=2019+4706=6725$.
Question 4. Calculate
$$
\log _{2}\left(\prod_{a=0}^{2018} \prod_{b=0}^{2018}\left(1+e^{\frac{2 \pi n k b}{2017}}\right)\right)
$$