Step 1: Recall the logarithmic forms.
We use $\sinh^{-1}t=\ln\left(t+\sqrt{t^2+1}\right)$ and $\cosh^{-1}t=\ln\left(t+\sqrt{t^2-1}\right)$, so $e^{\sinh^{-1}t}=t+\sqrt{t^2+1}$ and $e^{\cosh^{-1}t}=t+\sqrt{t^2-1}$.
Step 2: Evaluate the first exponential.
With $t=2\sqrt2$, $t^2=8$, so $e^{\sinh^{-1}(2\sqrt2)}=2\sqrt2+\sqrt{8+1}=2\sqrt2+3$.
Step 3: Evaluate the second exponential.
With $t=3$, $t^2=9$, so $e^{\cosh^{-1}(3)}=3+\sqrt{9-1}=3+2\sqrt2$.
Step 4: Add the two results.
Their sum is $(2\sqrt2+3)+(3+2\sqrt2)=6+4\sqrt2$.
Step 5: Simplify the target option.
For option (3), $\operatorname{Sech}^{-1}\left(\frac13\right)=\cosh^{-1}(3)$ because sech and cosh are reciprocals, so $e^{\operatorname{Sech}^{-1}(1/3)}=3+2\sqrt2$, and $2e^{\operatorname{Sech}^{-1}(1/3)}=2(3+2\sqrt2)=6+4\sqrt2$.
Step 6: Match.
This equals our sum exactly, so option (3) is correct.
\[ \boxed{2e^{\operatorname{Sech}^{-1}\left(\frac13\right)}} \]