Mixing different batteries, whether in parallel or in series, is a bad idea. Even enough of the same type of battery wired together is a bad idea.
Batteries are complicated electro-chemical devices. They vary quite a bit with temperature, age, state of charge, discharge rate, and what your dog had for breakfast. There is enough variation between "identical" batteries coming off the same assembly line that you have to be careful in combining them. With different batteries, the answer is simply "don't do that".
With primary cells, there is more forgiveness since they are used once until dead, and therefore future ability to be rechargeable and hold a predictable amount of energy is irrelevant. As such, it is allowable to let primary cells possibly suffer some damage as you try to get the last bit of energy from them. However, if you go too far, they could rupture and cause physical damage to whatever equipment they are contained in.
Rechargeable cells have to be treated much more carefully. Even just putting 4 cells, for example, in series is not so simple. One cell will inevitably have lower capacity than the others. This cell could be discharged so far as to cause damage, even when the stack voltage looks OK (the average cell voltage is above the damage limit).
Charging has the same issue. The lowest capacity cell can be overcharged and damaged before the highest capacity cell is really full.
Properly designed systems with multiple rechargeable cells in series will have monitoring of individual cells, and usually some kind of "charge balancing" circuitry. This shunts some of the charge current around the higher voltage cells to let the lower voltage cells catch up. On discharge, the current is stopped as soon as the lowest cell gets to the point where continuing would cause damage.
To make things more tricky, there isn't a hard line between damage and no damage. There are gray areas where deeper discharge or higher charge starts to decrease lifetime of the cell. That lifetime is itself really a probability.
It gets complicated, and proper multi-cell battery management is a deep topic and will get significant design attention in real professional systems intended for a long life.
Added in response to comment
The kanine nurishment dependency can have particularly tricky influence on performance, so has attracted much research. According to a study commissioned by the Project for Object Oriented Paradigms and performed at the Boise Institute for Technical Excellence, this complex relationship can be distilled down to one simple to understand graph:
This one:
"the current supplied remain constant and the batteries just drain less"
The LED current will be unaffected by the addition of the second identical parallel battery.
V = I x R
In this circuit you are doubling the battery, but not changing the output voltage (two identical 9V batteries in parallel is still a 9V output).
On the load side, the resistor and LED have not changed (that's the R in Ohm's law). Please note an LED is not accurately modeled as a pure resistance, but a complete explanation of that is not necessary to understand the answer to your question.
No change in V; No change in R; ...therefore NO CHANGE in I (current)
E = V x I x t
What does change is the total potential energy in this circuit. If you double the battery count, the total current sourced to the LED will be unchanged, but the current supplied by each battery will be 1/2 of the total. Because the batteries are supplying half the current as before, they will last twice as long.
Energy is voltage times current times the time the current is supplied at that voltage. A 1000mAh Alkaline battery means that it can supply 1A at ~1.4V for ~1 hour.
So...
No change in E; No change in V; ...therefore battery life (time) is INVERSELY proportional to current
Best Answer
First of all, these batteries are not ideal!
Convert both voltage sources (with thier resistor) to the current current source equivalent. The left one will be a 1A-source with a 1 Ohm resistor in parallel, the right one will be a 1A-soure with a 2 Ohm resistor in parallel.
Then you can add both current values to 2A. With the 0.66 Ohm resistor you calculated, the voltage will be 2A * 0.66 Ohm = 1,33 V.
Sounds pretty much like homework...