Here's The Answer:
You should always be looking to enter a planetary system such that you're travelling counter-clockwise when viewing the north pole of the parent body, since this is the "correct" direction and nothing orbits in the "wrong" direction.
When encountering a moon that is tidally locked, it effectively doesn't matter which direction you enter from, but since you might not know if a moon is tidally locked, and since nothing rotates slower than the tidally locked rotation speed, your best bet is to also enter in the "correct" direction. There is a small cost associated with landing or taking off in the retrograde direction, the worst being for Laythe, where the difference is only about 120m/s.
When launching from a planet or moon, launch towards the direction that the background stars rise. In the general case, this is usually the same direction that a parent star or planet rises from, and in KSP, is always the case.
Here's Why:
I'm going to get a bit sciency here, so please forgive that. It's necessary to properly explain what it means for a planet to "rotate the wrong way".
First, let's consider a two body system where the two bodies have masses m1 and m2 and the mass ratio m1 : m2 is greater than 1000:1, i.e. we have a star and an orbiting planet. To simplify things (and to mimic how KSP does things), we'll assume the star remains stationary (except possibly for rotation), effectively meaning that the barycenter of the system (the centre of mass of the system) is at the centre of mass of the star. This doesn't represent reality, but neither does KSP.
From this system, we can extract a number of angular momentum vectors: each body has one located at their centre of mass and extending along the axis of rotation towards the north pole (the rotational angular momentum vector), as well as a second one located at the barycentre of the system and pointed along the normal of the orbital plane (the orbital angular momentum vector, which also defines the ecliptic plane of a planet). Note that the first set of vectors is how north is defined for a body.[See note below] It is independent of the direction a planet or moon orbits the parent body, but I'll come back to this in a minute.
The first set of vectors are approximately 2/5mR2ω where m is the mass of the body, R is the radius, and ω is the angular velocity, while the second set are approximately m(r x v), i.e. the cross product of orbital distance and orbital velocity multiplied by the mass. In our simplified case, the orbital angular momentum vector for the star is zero because the orbital distance is zero. Additionally, we can define another angular momentum vector which is simply the vector sum of the previous angular momentum vectors, and defines the invariable plane. It also defines what it means to be "pointing north" relative to the system as a whole.
Now that we have these definitions, it's very easy to expand the number of children bodies, or even include whole sub-systems.
What does this mean for your question? Well a planet or moon only rotates "in the wrong direction" if the angle formed by the rotational angular momentum vector and the orbital angular momentum vector are more than 90°. This angle is defined as the axial tilt. No body in KSP meets this definition. (One possible exception would be Bop. I haven't personally checked it, and the wiki is a bit unclear.) In reality two planets in the solar system meets this definition: Uranus, though only just, with an axial tilt of almost 98°, and Venus, with an axial tilt in excess of 177°.[See note below]
It's also possible to orbit "in the wrong direction", by a similar definition, if the angle formed by the orbital angular momentum vector and the system angular momentum vector (that defines the invariable plane of the system) is greater than 90°. No body in KSP meets this definition. In reality, this is practically impossible for any planetary body that formed in the accretion disk of the young solar system.
As for tidally locked moons, yes there are plenty of them in KSP. Duna is even co-locked to its moon Ike. They are, however, still rotating in the "correct" direction. Even if these moons rotated slightly slower, they would still be rotating in the "correct" direction, since from an inertial frame of reference, they are still rotating in the same direction that they're orbiting. Yes, from a rotating frame of reference, they may appear to be rotating in the wrong direction, but we're not allowed to use a rotating frame of reference. It doesn't matter which direction a parent body rises from, although this is somewhat of a moot point for KSP since it doesn't happen.
Note:
The IAU doesn't actually use the mathematical definition described above for determining which pole is the North Pole, despite the convenience the mathematical definition affords when doing calculations. They instead prefer to designate the pole that forms the smallest angle with the invariable plane normal as the North Pole. From a physics and engineering standpoint, this is stupid, and may actually cause the definition of north on Uranus to eventually swap due to axial procession.
Eve has an atmosphere ranging all the way up to around 100,000m with surface pressure being about five times as high as on Kerbin, and almost twice as much gravity (with 16.7m/s2 gravitational pull).
Depending on the height from which you're launching, you will need between ~8,000m/s and ~11,500m/s of Δv to achieve orbital velocity around Eve, which should give an orbital velocity of ~3,500m/s.
Eve's escape velocity is ~5,000m/s, so in order to return from a 100km Eve orbit to Kerbin, another ~1,500m/s Δv is required.
Best Answer
You can go back to KSC and look in the Astronaut Complex. Kerbals on a mission are listed there, too.