I prefer to use the language of algebraic groups.
All algebraic groups and Lie algebras are defined over $\Bbb C$.

**1.** Let ${\mathfrak g}$ be a semisimple Lie algebra.
Consider the automorphism group ${\rm Aut\,}{\mathfrak g}$,
its identity component $G^{\rm ad}:=({\rm Aut\,}{\mathfrak g})^0$,
and the *group of outer automorphisms* ${\rm Out\,} {\mathfrak g}:=({\rm Aut\,} {\mathfrak g})/({\rm Aut\,} {\mathfrak g})^0$.
We say that $G^{\rm ad}$ is the *adjoint group* (or the group of adjoint type) with Lie algebra ${\mathfrak g}$. Note that $Z(G^{\rm ad})=\{1\}$.

**2.** Starting with a semisimple Lie algebra ${\mathfrak g}$,
one can construct the *simply connected group* $G^{\rm sc}$ with Lie algebra ${\mathfrak g}$;
see Steinberg, Lectures on Chevalley groups, AMS, 2016.
Note that $\pi_1(G^{\rm sc})=\{1\}$.
This algebraic group $G^{\rm sc}$ has the following universal property:
for any algebraic group $H$ with Lie algebra ${\mathfrak h}$
and for any homomorphism of Lie algebras $\varphi_{\rm Lie}\colon {\mathfrak g}\to{\mathfrak h}$,
there exists a unique homomorphism of algebraic group $\varphi\colon G^{\rm sc}\to H$ inducing $\varphi_{\rm Lie}$.

**3.** For any connected algebraic group $G$ with Lie algebra ${\mathfrak g}$, there exists a canonical surjective homomorphism
$$\rho\colon G^{\rm sc}\to G $$
inducing the identity isomorphism on ${\mathfrak g}$; see above.
We have
$$\pi_1(G^{\rm sc})=\{1\},\quad \pi_1(G)={\rm ker}\,\rho.$$
On the other hand, we have a canonical surjective homomorphism
$${\rm Ad}\colon G\to G^{\rm ad}\subseteq {\rm Aut\,} {\mathfrak g}$$
with kernel $Z(G)$.
Write
$$C=Z(G^{\rm sc})=\pi_1(G^{\rm ad}).$$
The homomorphism
$$ {\rm Ad}\colon G\to G^{\rm ad}$$
induces a homomorphism
$$i\colon \pi_1(G)\to\pi_1(G^{\rm ad})=C.$$
Moreover, the homomorphism
$$\rho\colon G^{\rm sc}\to G$$
induces a homomorphism
$$j\colon C=Z(G^{\rm sc})\to Z(G).$$
In this way we obtain a short exact sequence
$$1\to\pi_1(G)\overset{i}{\longrightarrow} C\overset{j}{\longrightarrow} Z(G)\to 1.$$

Conversely, for each subgroup $F\subseteq C$ one can associate a connected semisimple group
$ G_F:=G^{\rm sc}/F$ with Lie algebra ${\mathfrak g}$, with fundamental group $\pi_1(G_F)=F$, and with center $Z(G_F)=C/F$.
In this way we obtain a canonical bijection between the set of subgroups of $C$ up to conjugation by ${\rm Out\,} {\mathfrak g}$
and the set of isomorphism classes of connected semisimple algebraic groups with Lie algebra ${\mathfrak g}$.
It is known that ${\rm Out\,} {\mathfrak g}$ is canonically isomorphic to ${\rm Aut\,} {\rm Dyn}({\mathfrak g})$,
where ${\rm Dyn}({\mathfrak g})$ is the canonical Dynkin diagram of ${\mathfrak g}$.

**4.** Let us return to our exceptional simple Lie algebras.
The group $C=C({\mathfrak g})$ can be found, for instance, in tables in the book by Bourbaki "Lie Groups and Lie Algebras, Chapters 4-6",
or in the book by Onishchik and Vinberg "Lie Groups and Algebraic Groups", Springer-Verlag, 1990.

For ${\mathfrak g}_2$, ${\mathfrak f}_4$, and ${\mathfrak e}_8$ we have $C({\mathfrak g})=\{1\}$.
Thus there is only one (up to isomorphism) algebraic group $G^{\rm sc}({\mathfrak g})=G^{\rm ad}({\mathfrak g})$ with Lie algebra ${\mathfrak g}$.

For ${\mathfrak g}={\mathfrak e}_6$ we have $C({\mathfrak g})\simeq {\Bbb Z}/3{\Bbb Z}$. This group has no nontrivial subgroups.
Thus there are exactly two connected algebraic groups (up to isomorphism) $E_6^{\rm sc}$ and $E_6^{\rm ad}$ with Lie algebra ${\mathfrak e}_6$.
We have
$$Z(E_6^{\rm sc})=\pi_1(E_6^{\rm ad})\simeq{\Bbb Z}/3{\Bbb Z}.$$

For ${\mathfrak g}={\mathfrak e}_7$ we have $C({\mathfrak g})\simeq {\Bbb Z}/2{\Bbb Z}$. This group has no nontrivial subgroups.
Thus there are exactly two connected algebraic groups (up to isomorphism) $E_7^{\rm sc}$ and $E_7^{\rm ad}$ with Lie algebra ${\mathfrak e}_7$.
We have
$$Z(E_7^{\rm sc})=\pi_1(E_7^{\rm ad})\simeq{\Bbb Z}/2{\Bbb Z}.$$

**5.** The real forms of a connected algebraic group of an exceptional type correspond bijectively to the real forms of
(or real structures on) its Lie algebra.
My favorite way to classify those is via Kac diagrams.
See Table 7 in the book by Onishchik and Vinberg.
The number of real forms is 2 for ${\mathfrak g}_2$, 3 for ${\mathfrak f}_4$, 3 for ${\mathfrak e}_8$, 4 for ${\mathfrak e}_7$, 5 for ${\mathfrak e}_6$.
These real forms are listed also in Table V in Chapter X of Helgason's book "Differential Geometry, Lie Groups, and Symmetric Spaces"
(Helgason lists all *non-compact* forms).
Helgason classifies real forms using the original method of Kac with infinite dimensional Lie algebras.
Onishchik and Vinbeg use another method, which gives exactly the same answer (the same Kac diagrams).

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