As you state, conformal time is defined as
$$
\eta(t) = \int_0^t\frac{\text{d}t'}{a(t')}.
$$
Using
$$
\dot{a} = \frac{\text{d}a}{\text{d}t},
$$
this can be written in the form
$$
\eta(a) = \int_0^a\frac{\text{d}a}{a\dot{a}} = \int_0^a\frac{\text{d}a}{a^2H(a)},
$$
with
$$
H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}.
$$
The scale factor $a$ is related to the redshift as
$$
1 + z = \frac{1}{a},
$$
so that
$$
\eta(z) = \frac{1}{H_0}\int_0^{1/(1+z)}\frac{\text{d}a}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a + \Omega_{K,0}\,a^2 + \Omega_{\Lambda,0}\,a^4}}.
$$
Basically, the conformal time is equal to the distance of the particle horizon, divided by $c$ (see this post for more info). $\eta_0$ refers to the current conformal age of the universe.
edit
I just checked the values that you posted with my own cosmology calculator. I get
$$
\eta_0 = 45.93\;\text{Gigayears}
$$
for the current conformal age of the universe, and
$$
\begin{align}
z_e &= 3234,& a_e &= 0.000309,& t_e &= 5.54\times 10^{-5}\;\text{Gy},&\eta_e &= 0.3804\;\text{Gy},\\
z_E &= 0.39,& a_E &= 0.719,& t_E &= 9.359\;\text{Gy},& \eta_E &= 41.08\;\text{Gy},\\
z_d &= 1089,& a_d &= 0000917,& t_d &= 0.00037\;\text{Gy},& \eta_d &= 0.911\;\text{Gy},
\end{align}
$$
so that
$$
\frac{\eta_e}{\eta_0} = 0.00828,\quad \frac{\eta_E}{\eta_0} = 0.894,\quad \frac{\eta_d}{\eta_0} = 0.0198.
$$
So my results are almost the same, but there's a small discrepancy. Apparently there's a small numerical error somewhere.
I am not sure this is what you want but I want to give it a try,
$$\eta=\int \frac {dt} {a}=\int\frac {da} {a\dot{a}}=\int\frac {da} {a^2H}$$ and we can write
$$H(z)=H_0E(z)$$
$$E(z)=\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3+\Omega_r(1+z)^4+\Omega_{\kappa}(1+z)^2}$$
so we have,
$$\eta=\int\frac {da} {a^2H_0E(z)}$$
and $dz=-da/a^2$ so we can write,
$$\eta=-H_0^{-1}\int\frac {dz} {E(z)}$$
And by taking initial coniditon as $z=\infty$, and due to the minus sign the integral becomes,
$$\eta=H_0^{-1}\int_z^{\infty}\frac {dz} {E(z)}$$
To find the current($t_0$) conformal time, we can use the above equation, for $z=0$
$$\eta=H_0^{-1}\int_{z=0}^{\infty}\frac {dz} {E(z)}$$
$$\eta=H_0^{-1}\int_{0}^{\infty}\frac {dz} {\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3+\Omega_r(1+z)^4+\Omega_{\kappa}(1+z)^2}}$$
For the current values of $\Omega_{\Lambda}=0.69$, $\Omega_m=0.31$,$\Omega_{\kappa}= \Omega_r=0$
we have,
$$\eta=H_0^{-1}\int_{0}^{\infty}\frac {dz} {\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3}}$$
$$\eta=H_0^{-1}\int_{0}^{\infty}\frac {dz} {\sqrt{0.69+0.31(1+z)^3}}$$
If we take $H_0=70km/s/Mpc$ then $1/H_0=1/(70\times 3.2408\,10^{-20})=4.4133353\,10^{17}s$
And the integral gives, $$\int_{0}^{\infty}\frac {dz} {\sqrt{0.69+0.31(1+z)^3}}=3.266054427285631$$
so $$\eta(t_0)=3.266054427285631\times 4.4133353\,10^{17}s=1.4414193\,10^{18}=45.70 \,\text {Gigayear}$$
To calculate the integral you can use, this site
I write the integral in terms of $z$ but, its also possible to write the equation in terms of $a(t)$ (the begining part of the derivation). But $z$ is the observable value so I prefer to write in that form.
For a given $t$ you can turn easily $a(t)$ to $z$.
Best Answer
These are essentially the same, as indicated by the formulas, with one having units of length and the other of time. So, (1) yes, the conformal time of emission of the light we see is simply today's conformal time minus ($1/c$ times) the comoving distance to the galaxy (though this is not directly measurable but rather must be found with e.g. the luminosity distance or the angular diameter distance).
(2/3) $\eta$ is a rescaling of "normal" time $t$ (proper time for an observer moving with the Hubble flow in an FRW universe), and so it still measures a "time." If you foliate spacetime with spacelike slices of constant time, then $\eta$ can measure the distance between the slices containing two different events. $\chi$ you can take to be the proper distance between two events, if both events are projected onto our current time slice by following the Hubble flow.
Note that the purpose of conformal time and comoving distance is to make the Robertson-Walker metric conformally equivalent to Minkowski (or something very similar in the non-flat case): $ds^2 \propto -c^2d\eta^2 + d\chi^2 + S_k(\chi)^2 d\Omega^2$. Thus photons ($ds^2 = 0$) traveling in the radial ($d\Omega = 0$) direction simply move a comoving distance $\Delta\chi = c \Delta\eta$ in a conformal time $\Delta\eta$, giving the simple relation between comoving distance and conformal lookback time applicable to (1).