The Short Answer
How is the efficiency of a heat engine related to the entropy produced during the process?
The maximum efficiency for any heat engine operating between two temperature $T_H$ and $T_C$ is the Carnot efficiency, given by
$$e_C = 1 -\frac{T_C}{T_H}.$$
Such a heat engine produces no entropy, because we can show that the entropy lost by the hot reservoir is exactly equal to the entropy gain of the cold reservoir, and of course, the system's entropy on the net doesn't change because the system undergoes a cycle.
Any heat engine operating between the same two temperatures whose efficiency is less than $e_C$ necessarily increases the entropy of the universe; in particular, the total entropy of the reservoirs must increase. This increase in entropy of the reservoirs is called entropy generation.
Finally, the efficiency of the perfect engine is less than one, necessarily, because the entropy "flow" into the system from the hot reservoir must be at least exactly balanced by the entropy "flow" out of the system into the cold reservoir (because the net change in system entropy must be zero in the cycle), and this necessitates waste heat from the system into the cold reservoir. The fact that $e_C$ goes to one in the limit of small ratios $T_C/T_H$ is a consequence of the fact that $Q_C$ is small compared to $Q_H$. It is not a consequence of the fact that entropy generation is small in this case, because entropy generation is already zero for the Carnot cycle.
Explanation
Let's concentrate first on the interaction between the system and the hot reservoir. An amount $\delta Q_H$ of energy flows into the system from the hot reservoir, which means that the system's entropy changes by
$$\mathrm dS_\text{sys} = \frac{\delta Q_H}{T_\text{sys}},$$
and correspondingly, the reservoir's entropy changes by
$$\mathrm dS_\text{hot} = -\frac{\delta Q_H}{T_{H}}.$$
It is straight-forward to show then, that the total change in entropy of system plus environment satisfies
$$\mathrm dS = \mathrm dS_\text{hot}+\mathrm dS_\text{sys} \geq0,$$
with equality holding if and only if the system and environment exchange energy via heating when they have equal temperatures, $T_\text{sys} = T_H$.
As a consequence, in order to minimize entropy production (and, in fact zero it out completely) during this process, we want $T_\text{sys} = T_H$, and the net change in system entropy during this process can then be written as
$$\Delta S_\text{sys} = \int \frac{\delta Q_H}{T_\text{sys}} = \frac{Q_H}{T_{H}},$$
since we are assuming that the temperature of the reservoir doesn't change at all during the cycle.
Now, since the system operates on a thermodynamic cycle, and since the system entropy $S_\text{sys}$ is a state variable (state function/$dS$ is an exact differential, etc.), it must be true that
$$\mathrm dS_\text{sys,cycle}=0.$$
Therefore, there must be some other process during which the system expels an amount of energy $Q_C$ to some other reservoir via heating in such a way that the change in system entropy during this new process is the negative of the change in system entropy that we calculated before. By the same argument as above, it must be that this change in entropy is
$$\Delta S_2 = -\frac{Q_C}{T_C},$$
where $T_C$ is the temperature of the cold reservoir.
Finally, then, since system entropy is a state variable,
$$0 = \Delta S + \Delta S_2 = \frac{Q_H}{T_H}-\frac{Q_C}{T_C}.$$
Another way of looking at this equation is that the net change in entropy of the hot reservoir is negative the net change in entropy of the cold reservoir during the cycle, and hence the net change in entropy of the universe is zero during the cycle.
Efficiency and work
Now, none of this seemed related to the fact that efficiency goes to 1 as the ratio of $T_C$ to $T_H$ goes to zero. This comes in in the following way. First, the net work output during one cycle is
$$W_\text{out} = Q_H-Q_C,$$
and hence the efficiency of the engine that we've just made is
$$e = \frac{W_\text{out}}{Q_H} = 1 - \frac{T_C}{T_H},$$
after some algebra. Based on our calculation above, this must be the maximum efficiency of any engine operating between these two temperatures. However, if we change the temperatures, then we can change the efficiency. The reason the efficiency goes up as the temperature ratio goes down is that $W_\text{out}$, being the difference between the heat flows, must go up if, say, we lower $T_C$ (because then $Q_C$ goes down) or if we raise $T_H$ (because then $Q_H$ goes up).
In some sense, this part really doesn't have much to do with entropy at all, because from the thermodynamic perspective, entropy production (which is the increase in entropy of an isolated system) is a measure of how much work we could have done if we had done the process reversibly, but we have already designed the perfect engine operating between those two particular temperatures above, so entropy doesn't have anything else to say.
Best Answer
Yes, if it is a reversible isothermal process, such as the reversible isothermal expansion of an ideal gas in the the Carnot cycle.
As already indicated, it can if it is a reversible isothermal process invoolving an ideal gas. However no real heat transfer process is reversible (or as you say, "practically possible") since heat transfer by definition requires a temperature difference (thermal disequilibrium). Another impracticality, that @David White pointed out in his comment below, the smaller the temperature difference the slower the heat transfer rate and the process will take an infinite amount of time. That severely limits the practicality of the cycle in terms of power output (work per unit time). In a reversible process the temperature difference is infinitesimally small so that we can say, in the limit, the process is 100% efficient and would not violate the second law, but the process would go extremely slowly. See further considerations discussed below regarding your Q4.
A process can be irreversible yet still be isothermal. Isothermal refers to the system temperature being constant which is possible if the system can be considered a thermal reservoir, that is, its heat capacity is sufficiently large that the amount of heat transferred into or out of the system does not change its temperature.
The difference between a reversible and irreversible isothermal process is in the former the system temperature is in equilibrium with the temperature of the surroundings (temperature difference being infinitesimally small) at all times, whereas in the case of the latter there is a finite temperature difference. The ramifications of this are discussed below an response to question 4.
An irreversible process restricts the conversion. An irreversible process generates entropy which, for a complete cycle, must be dumped back into the surroundings in order to return all system properties (including entropy) to their original state. The only way to transfer entropy to the surround is by transferring some of the heat the system got during the expansion back to the surroundings during the isothermal compression process. That heat is thus unavailable to produce net work in the cycle. To further explain:
For a reversible isothermal heat transfer (where the system and surroundings temperature is the same) the change in entropy of the system is
$$\Delta S_{sys}=+\frac{Q}{T}$$
And the change in entropy of the surroundings is
$$\Delta S_{surr}=-\frac{Q}{T}$$
For a total entropy change (system + surroundings) of
$$\Delta S_{tot}=\Delta S_{sys}+\Delta S_{surr}=+\frac{Q}{T}-\frac{Q}{T}=0$$
This satisfies the equality $\Delta S_{tot}=$ 0 for a reversible cycle.
For an irreversible isothermal process the temperature of the surroundings is greater than the system. So the change in entropy of the surroundings is
$$\Delta S_{surr}=-\frac{Q}{T_{surr}}$$
And the system
$$\Delta S_{sys}=+\frac{Q}{T_{surr}-\Delta T}$$
The total entropy change then being
$$\Delta S_{tot}=+\frac{Q}{T_{surr}-\Delta T}-\frac{Q}{T_{surr}}$$
Note that for all positive values of $\Delta T$ the first term will be greater than the second and therefore, for an irreversible heat transfer we have
$$\Delta S_{tot}>0$$
Since all real processes are irreversible, the inequality holds for all real processes.
The difference between the change in entropy of the system and the surroundings is call the entropy generated in the system. To complete a cycle and return all properties to their original values this entropy generated must be returned to the surroundings which can only occur in the form of heat. That heat is thus not available to do work in the cycle. So the net work done in an irreversible cycle will always be less than that for the same cycle carried out reversibly.
In short, for an irreversible isothermal expansion not all of the heat transferred to the system produces net work in the cycle because of the entropy generated. Some has to be returned to the surroundings.
Hope this helps.