Hermitian operators in the expansion of symmetry operators in Weinberg's QFT

This is related to Taylor series for unitary operator in Weinberg and Weinberg derivation of Lie Algebra.

$textbf{The first question}$

On page 54 of Weinberg’s QFT I, he says that an element $T(theta)$ of a connected Lie group can be represented by a unitary operator $U(T(theta))$ acting on the physical Hilbert space. Near the identity, he says that
$$U(T(theta)) = 1 + itheta^a t_a + frac{1}{2}theta^atheta^bt_{ab} + ldots. tag{2.2.17}$$
Weinberg then states that $t_a$, $t_{ab}$, … are Hermitian. I can see why $t_a$ must be by expanding to order $mathcal{O}(theta)$ and invoking unitarity. However, expanding to $mathcal{O}(theta^2)$ gives
$$t_at_b = -frac{1}{2}(t_{ab} + t^dagger_{ab})tag{2},$$
so it seems the same reasoning cannot be used to show that $t_{ab}$ is Hermitian. Why, then, is it?

$textbf{The second question}$

In the derivation of the Lie algebra in the first volume of Quantum Theory of Fileds by Weinberg, it is assumed that the operator $U(T(theta)))$ in equation (2.2.17) is unitary, and the rhs of the expansion
begin{equation}
U(T(theta)))=1+itheta^a t_a +frac{1}{2} theta_btheta_c t_{bc} + dots
end{equation}
requires
$$t_{bc}=-frac{1}{2}[t_b,t_c]_+.$$
If this is the case there is a redundancy somewhere. In fact, by symmetry
$$
U(T(theta))=1+itheta_at_a+frac{1}{2}theta_atheta_bt_{ab}+dotsequiv 1+itheta_at_a-frac{1}{2}theta_atheta_bt_at_b+dots
$$
and it coincides with the second order expansion of $expleft(itheta_at_aright)$; the same argument would then hold at any order, obtaining $$U(T(theta))=expleft(itheta_at_aright)$$ automatically.
However, according to eq. (2.2.26) of Weinberg’s book, the expansion
$$U(T(theta))=expleft(itheta_at_aright)$$
holds only (if the group is just connected) for abelian groups.
This seems very sloppy and I think that Lie algebras relations could be obtained in a rigorous, self-consistent way only recurring to Differential Geometry methods.

There have been some answers or speculations for these two questions but I do not think they are solved. I think the crucial point for these two questions is that the $t_{ab}$ operator is $textbf{not}$ hermitian unless the ${t_a}$ operators commute with each other. Here is why:

From the unitarity of $U(t(theta))$ we have
$$t_at_b = -frac{1}{2}(t_{ab} + t^dagger_{ab})tag{2},$$
And from the expansion of $f(theta_a,theta_b)$ we have
$$t_{ab} = t_a t_b – if^c_{ab} t_c.$$
So $t_{ab}$ is hermitian iff $f^c_{ab}$ is zero, which mean that ${t_a}$ group algebra is abelian.

I think that solves the problem. Any other opinions?