In the 1st page of the introduction of Hazewinkel's Formal groups and Applications book, there are two ways of constructing formal groups (law):

$\bullet$ Given a Lie group $G$, one can define a formal group $F$ around the identity $e \in G$. Further, one can associate a Lie algebra $\mathfrak{g}$ to every formal group $F$. That the formal group $F$ is a bridege between the Lie group $G$ and Lie algebra $\mathfrak{g}$.

$\bullet$ Given a L-function $L(s)=\sum_{n=1}^{\infty} \frac{a(n)}{n^s},~a(n) \in \mathbb Z$, one can associate a formal power series $f_L(X)=\sum_{n=1}^{\infty}\frac{a(n)}{n}X^n \in \mathbb Q[[X]]$ and hence a formal group $F_L(X,Y)=f_L^{-1}(f_L(X)+f_L(Y))$.

Then Hazewinkel mentions that the formal groups $F$ and $F_L$ are not independent, that is, there is a relation between them.

$(1)$ What is the relation of $F$ and $F_L$? For arbitrary L-function, do we know the relation ?

As an example, Hazewinkel mentions: formal group (formal compltion) of a elliptic curve $E$ over $\mathbb Q$ gives some beautiful results concerning the zeta function of $E$. Regarding this, there is a famous work by T. Honda in the paper Formal groups and zeta functions, Osaka J. Math. 5 (1968), 199-213.. For,

let $d$ be a discriminant of a quadratic number field $K=\mathbb Q(\sqrt d)$ and let $L'(s)=\sum_{n=1}^{\infty} \left(\frac{d}{n}\right)X^n$ be a Dirichlet L-function, where $\left(\frac{d}{n}\right)$ is Kronecker sysmbol. Then Honda says, the formal group associated to $L'$ is isomorphic to $F(X,Y)=X+Y+\sqrt d XY$ over the ring of integers of $K$.

$(2)$ What are the unsolved problems along this direction ?

$(3)$ Is there any generalization of Honda's work beyond quadratic extension of $\mathbb Q$? e.g., for cubic or any finite abelian extension

More specification, suppose we are an arbitrary L-function, do we know the associated formal group (law)?

Note: I have been studied about formal groups over $p$-adc number field in my PhD works and so I want to continue my investigation in formal groups.

Your suggestion/guidance is much appreciated.