Unfortunately, this is not true in general.
Let $E$, $F$, $G$ and $H$ be complete nuclear lcs such that
$$ 0 \longrightarrow E \stackrel{S}{\longrightarrow} F \stackrel{T}{\longrightarrow} G \longrightarrow 0 $$
is a strict short exact sequence. Note that the sequence
$$ 0 \longrightarrow E \mathbin{\hat\otimes} H \stackrel{S \mathbin{\hat\otimes} \text{id}_H}{\longrightarrow} F \mathbin{\hat\otimes} H \stackrel{T \mathbin{\hat\otimes} \text{id}_H}{\longrightarrow} G \mathbin{\hat\otimes} H \longrightarrow 0 $$
is a strict short exact sequence if and only if $T \mathbin{\hat\otimes} \text{id}_H : F \mathbin{\hat\otimes} H \to G \mathbin{\hat\otimes} H$ is surjective; everything else is automatic by the mapping properties of the injective resp. projective topology.
Let $w$ and $w*$ denote the weak and weak-$*$ topologies, and let $T' : G_{w*}' \to F_{w*}'$ denote the adjoint of $T$.
It follows from [Sch99, §IV.9.4] that $F \mathbin{\hat\otimes} H \cong \mathfrak{B}_e(F_{w*} \times H_{w*})$. The latter is linearly isomorphic with $\mathfrak{L}(F_{w*}' , H_w)$, and the following diagram commutes:
$$\require{AMScd}
\begin{CD}
F \mathbin{\hat\otimes} H @>T \mathbin{\hat\otimes} \text{id}_H >> G \mathbin{\hat\otimes} H \\
@VV V @VV V \\
\mathfrak{L}(F_{w*}' , H_w) @> R \mapsto RT' >> \mathfrak{L}(G_{w*}' , H_w) \\
\end{CD}$$
Since $G \cong F/E$, we have $G' = E^\perp$, and the weak-$*$ topology on $G'$ coincides with the relative $F_{w*}'$ topology of $E^\perp$. Thus, we see that $T \mathbin{\hat\otimes} \text{id}_H$ is surjective if and only if every weak-$*$-to-weak continuous operator $E^\perp \to H$ can be extended to $F'$. This is not always the case:
Counterexample. Let $E$ be a closed, non-complemented subspace of a nuclear Fréchet space $F$.¹ Furthermore, let $G := F/E$, and let $H := G_\beta'$ be the strong dual of $G$. Then $E$, $F$ and $G$ are nuclear Fréchet spaces (so in particular reflexive), and $H$ is a complete, nuclear (DF)-space.
In a Fréchet space, all weakly complemented subspaces are complemented, so $E$ is not weakly complemented in $F$. By duality, $E^\perp = G'$ is not weak-$*$ complemented in $F'$. Since $G$ is reflexive, the weak and weak-$*$ topologies on $H = G_\beta'$ coincide, so we have $\text{id}_H \in \mathfrak{L}(G_{w*}' , H_w)$. Since $G'$ is not weak-$*$ complemented, the map $\text{id}_H \in \mathfrak{L}(G_{w*}' , H_w)$ has no extension in $\mathfrak{L}(F_{w*}' , H_w)$.
Convesely, if $E$ is (weakly) complemented, then $T \mathbin{\hat\otimes} \text{id}_H$ is always surjective.
¹: Concrete examples of this kind are given in this answer, or in [MV97, Exercise 31.4], or in [DM76], among others.
References.
[DM76] Plamen Djakov, Boris Mitiagin, Modified construction of nuclear Fréchet spaces without basis, Journal of Functional Analysis, vol. 23 (1976), issue 4, pp. 415–433. DOI: 10.1016/0022-1236(76)90066-5.
[MV97] Reinhold Meise, Dietmar Vogt, Introduction to Functional Analysis (1997), Oxford Graduate Texts in Mathematics, Clarendon Press, Oxford.
[Pie72] Albrecht Pietsch, Nuclear Locally Convex Spaces (1972), Ergebnisse der Mathematik und ihrer Grenzgebiete, Springer, Berlin.
[Sch99] H.H. Schaefer, M.P. Wolff (translator), Topological Vector Spaces, Second Edition (1999), Springer Graduate Texts in Mathematics, Springer, New York.
