sqrt(exp)

[Kouznetsov]

It seems we dont speak about the same thing.
So my first question is:
\( \sqrt{\exp}(L_{e,1})=L_{e,1} \)?
where \( L_{e,1}\approx 2.062 + 7.589i \) is the fixed point of \( \exp \) on the upper halfplane that has the second lowest distance to the real line.

My first answer is NO.
\( \sqrt{\exp}( 2.062277729598284 + 7.5886311784725127 ~\mathrm{i})
\approx -17.11069793592735 + 5.77820343698599 ~\mathrm{i}
\).
Looking at the graphic, I see, that the \( \Re\Big(\sqrt{\exp}(L_{\mathrm{e},1})\Big) \) is negative.

I mean one could expect that a half iterate has the same fixed points as the function itself.

No, One knows that if \( z^2=1 \), then it does not imply that \( z=1 \).

Unfortunately I can not verify the above question from the picture.

I can. It is difficult to see that the real part is of order of \( -17 \), but it is easy to see that it is negative. Do you want me to draw more levels for negative values of the real part of \( \sqrt{\exp} \)?

So if it turns out that \( \sqrt{\exp}(L_{e,1})\neq L_{e,1} \) on the given domain of definition, then I would expect that there is some branch\( \sqrt{\exp}_k \) such that \( \sqrt{\exp}_k(L_{e,1})=L_{e,1} \).

Yes, and you will have to determine somehow the positions of all new cutlines you create in such a way.

And further if there is such a branch (made up at the branch point \( L \)) then I would expect that \( \sqrt{\exp}_k \) has a (maybe isolated) singularity at \( L_{e,1} \).

I doubt about "isolated". There should be a branchpoint. If you want to keep \( \mathrm{slog}\big(z^*\big)=\mathrm{slog}(z)^* \), two additional branchpoints.
I expect, you will get a pair of new branchpoints per each turn. You may combine the cutlines at the real axis; then the \( \sqrt{\exp} \) will not be defined at the real axis.