The RED LED should not be glowing. Maybe something is wired wrong or the 2N7000 is damaged, but there should not be more than nA of current flowing with 10mV on the gate, which will not produce a visible amount of light in normal room lighting, even with the most efficient red LED. Try replacing Q2 after checking the wiring.
Of course if you are leaving some types of multimeter connected where you are measuring the 0.35V then there could be enough current through the meter to provide a faint light from the LED.
There is no problem in reverse biasing LEDs up to the rated voltage (usually 5V), and in fact far beyond that typically (but there are no guarantees so you shouldn't do that).
First off, your high-frequency flashing will give (at best) only a slight improvement in battery lifetime. The idea is that you can increase efficiency by driving the LED at a higher current, and while this is true for some LEDs, it's not true for all.
That said, getting a year out of a continuously flashing LED will take a rather large battery. Let's say that you're flashing the LED with a 25% duty cycle: that is, only on for 25% of the time. And furthermore, let's say that you are running the LED at 10 mA. Then (obviously) in one hour the LED will be on for 15 minutes. The total current drawn will be 10 mA x 0.25 hours, or 2.5 mA-hr. In one day, the total current drawn will be 2.5 x 24, or 60 mA-hr. After one month, it will be 60 x 30, or 1800 mA-hr.
Now for batteries. You can get rechargeable NIMH D cells that will give you 10 - 12 A-hrs of service, so you might think to get a year at lower current drains. However, http://data.energizer.com/PDFs/nickelmetalhydride_appman.pdf suggests that NIMH will self-discharge by 20 - 50% in 6 months, so that pretty much sets that limit.
A bog-standard alkaline D cell will have http://data.energizer.com/PDFs/e95.pdf a capacity of ~ 18 - 20 A-hr and in this application you'd think would give you about 10 months to a year. The problem here is that the lifetime is specified to the point that the battery voltage is 1/2 that of a fresh battery, and LEDs have a minimum voltage that they need.
So, what you want to do is "possible", but only if you can accept the battery size.
What you need to do is get some high-brightness LEDs, a cheap power supply, a DMM and some resistors, and experiment with the LEDs. Find the current level that gives you the brightness you need, keeping in mind that your installation may (or may not, you just have to think about it) be seen in bright light which will require more LED current to appear bright enough. Once you have a target current, only then can you start grinding out the numbers which will tell you how much battery you need.
For example, let's say you have 2 LEDs in series, and they draw 10 mA when on, and their forward voltage is 2 volts each. The LEDs will need 4 volts total. If you use alkaline D cells you can put 6 in series to get 9 volts, and at the end of life they will be putting out about 4.8 volts, giving you about 0.8 volts margin for your current control element. Assuming 20 A-hr for the cells, you'll get about 11 months.
Is that too big a battery? Well, there are other alternatives. For this long-duration application, http://batteryuniversity.com/learn/article/elevating_self_discharge indicates that rechargeable lithium cells (lithium-ion) are possible. http://www.electricwingman.com/power/a123-loose-cells.aspx has a single 20 A-hr cell. Note its size and weight.
Primary lithiums are also a good technical choice. http://www.onlybatteries.com/showitem.asp?ItemID=14509&vfpr=10.30&vfbr=Blue+Sleeve&vfcat=Electronics+Electronics+Accessories+Power+Batteries&vfsku=14509&vfbi=no_bid&sid=gpla&vfsku=14509&vfsku=14509&gpla=pla&gclid=CjwKEAjwuoOpBRCSy6yQm66J1g8SJABrXW48PiokBnHnnAVNcptv7FH19wDGumSYhbMnC9kQH9-IQxoCevPw_wcB is an example, and you could get away with one of them.
The point of that connection is to make the initial current through the LED high and then turn off, regardless of the low time of the signal. How and when that happens depends on the strength of the generated signal.
If the signal has been high a while, the voltage across the capacitor is very low, in the order of 0.2V or possibly less. (verify for yourself).
Then if the signal goes low, the LED will "see" its anode to be about 3V, so it will conduct what current belongs with 3V across it, as long as the signal can sink that current. The current flowing will charge the capacitor, so the voltage across the capacitor increases. This means the LED voltage decreases, which will make it conduct less and less current until it turns off.
Then when the signal goes high again the diode will very quickly source current into the capacitor the other way, to discharge it.
The time-base at which that happens can easily be a milisecond or even less, depending on the LED, so you might not be looking "fast enough", which is why you see the square wave. The last little slant in the high period is the residual current of the Schottky charging the capacitor a bit more. "Officially" the diode already stopped conducting, but below its Vf at 1mA it will still allow a a trickly current to flow, eventualy all the way down to single uA if you wait long enough.
So this seems like a trick to make a LED quickly pulse brightly on an otherwise slow signal used for something else. Or just to make it pulse brightly in the first place.