Welcome to WB-SE. I hope you shall find it a friendly place, and come to pronounce it "wubbsy".
There are a couple of unclear points on your question, so I'll make some assumptions: feel free to correct me.
1) When you say "facing towards" and "facing away", I presume that you mean only one hemisphere of each moon is habitable for some reason (perhaps the inward face of the outer moon and outer face of the inner moon are ravaged by tectonic tidal effects as they pass each other, making them uninhabitable?).
2) When you say "dusk" for moon 1 and "night" for moon 2, you mean "the sun is NOT directly visible from the inhabited surface", and when you say "day" you mean "the sun is directly visible."
3) When you say "axial tilt" I'm assuming you mean that the planet has an axial tilt relative to the sun that's approximately the same as the Earth's, and that the moons are orbiting the planet in that tilted plane: they do not have an axial tilt relative to the planet. That is, they're in the same plane as any debris disk and such. Which, as you say, gives the moons the same "year" as the planet itself.
4) I'm gonna call "north" the same direction for all entities (sun, planet, moon), and assume that, looking down on to their north poles, all are rotating counter-clockwise, as in our solar system. Reverse west and east in the following if that's incorrect.
5) You are asking a very specific question about how will the light from the sun will reach the habitable hemispheres of the two moons, as they orbit. While you welcome corrections to the values in your system, you have already calculated them and are not asking for help with them, nor with a "calendaring system" nor with "habitable zones", etc. [Edit: this assumption is wrong, see edit below.]
If all of the above are true, then I think you are essentially correct: anywhere on the hemisphere facing the planet, you will always see the planet, and the sun will appear to rise to the West, pass behind the planet when the moon is furthest from the sun (which is "midday" for the point on that hemisphere closest to the planet), and then set in the East, with the darkness extending either side of that for a time which depends on the width of the planet.
That eclipse-duration is unlikely to be a precise one-third of the orbit, but can be calculated, by finding what the moon's orbital radius and the planet's radius is. Back-of-an-envelope, I think the duration of the eclipse varies with:
MoonOrbitalPeriod * 360/arcsin(PlanetRadius/MoonOrbitRadius).
[Edit: I am skeptical of the above expression and will investigate further once I can obtain a better envelope upon which to doodle.]
At the point closest to the planet, the night will last roughly half the orbit, give or take for slight atmospheric lensing of the sun's light. However, on the rest of the hemisphere facing the planet, light before eclipse will be either shorter or longer than after. So, on the hemisphere facing the planet, the two extremes will be on the Western and Eastern edges. Let's look at each one.
To the far West, you will be facing away from the sun until halfway through the eclipse, and once emerging from the eclipse will be in daylight for (half an orbit - half an eclipse-time). The western edge will be the first to emerge from the eclipse, and sunset will be at the moment of closest approach to the sun. This will be "midnight" for the point closest to the planet, and will be dawn for the point farthest East.
That eastern point will remain lit until it enters the eclipse (it is on the trailing edge, and is the last to be covered by the planet's shadow).
For a moon's side facing away from the planet, everything is backwards (sun rises in the east, sets in the west, just like Earth, with longer days), and there're no eclipses.
Now let's take into account the axial tilt.
On the outer moon, with the habitable face away from the planet, the tilt's the same as Earth's near enough, so has the basic same effect of giving "seasons", where the north or south is lit more or less than its counterpart. To the farthest north and south, in their "winter" they may not see the sun for months; and in "summer", they may see it from months on end, possibly with interruptions from eclipses.
On the inner moon, facing the planet, this is the same, with the additional effect that this will also affect the ground-track of the centre-point of the eclipse.
This means that, to the far north and south, you may have no eclipses at all during their "summers", depending on the radius of the planet. Again back-of-an-envelope, but I think that in summer you'll see over the top of the planet to the sun at the very northern of pole the moon if the planet's radius is less than:
(sin(23) * LunarOrbitRadius) + LunarRadius
You'll see no eclipses anywhere on the moon in summer and winter if the planet's radius is less than:
(sin(23) * LunarOrbitRadius) - LunarRadius
As the year wears on, the planet will take a bigger and bigger bite out of the top or bottom of the sun, until again the eclipses are total during spring and fall.
Hope that helps with the lighting of your system.
Edit: I owe @JHB and @Cyn an apology: reading comprehension fail on my part, as the very first paragraph clearly asked multiple questions. Sorry guys!
I do feel that this was an interesting astrodynamics problem, and the question could trivially be split to ask about it, and the other questions asked, if spun off, could be interesting in themselves.
This is an edited hopefully more focused version of a question I asked a while ago. I made the mistake of asking too many questions the first time so I’ll try and be a bit more focused here. 
