= Solution
Set $S=A+B$ and $E(t)=F(t)-e^{tS}$. Since $F(0)=I$, $E(0)=0$. Differentiate the ordered exponential products, using that each <matrix> commutes with its own exponential:
$$
F'(t)-SF(t)
=\frac12\left\{[e^{tB},A]e^{tA}+[e^{tA},B]e^{tB}\right\}
=:D(t).
$$
Here the <commutator> convention is $[X,Y]=XY-YX$; this fixes both signs. The error satisfies $E'=SE+D$, so the <variation-of-constants formula> gives
$$
\boxed{E(t)=\frac12\int_0^t e^{(t-x)S}
\left\{[e^{xB},A]e^{xA}+[e^{xA},B]e^{xB}\right\}\,dx.}
$$
This is the <symmetrized exponential-splitting defect identity>. Symmetry of $A,B$ was not needed for the identity itself; it will be used to bound their exponentials in part (c).
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