(How) can we determine mid-point on Earth's orbit? The eccentricity of earth's orbit varies with time, but at present time its eccentricity is roughly mean e (0.0167).
The position of equinoxes is far more complicated than I thought and it is not at mid-point, can someone explain how you determine with a certain accuracy the middle point of the ellipse (where $\lambda = \pi/2 $ (or $3/2 \pi$)?
Edit: 


*

*is the formula given by Pulsar valid for any point of the ellipse (for example $\lambda = 2.4981 = 143.13°$ from perihelion? (143.13- 0.0167*.6?)

 A: 
The position of equinoxes is far more complicated than I thought, can someone explain how you determine with a certain accuracy the middle points of the ellipse B and C?

Your points B and C will not help in your understanding.

Point C was before the end of September, point B in March, how can we determine the position with a certain accuracy?

From your diagram, point B was on April 3 (more or less) and point C, October 3. These points are where the eccentric anomaly is 90 degrees (point B) and 270 degrees (point C). Unlike periapsis and apoapsis, which have lots of different names (e.g., perigee, perihelion, perilune, perijove, etc., depending on the body of interest), your points B and C have no name because there's nothing special about those points.
The timing of the equinoxes and solstices is fairly simple. In a non-leap year, an equinox or solstice will be about 5 hours and 49 minutes (plus or minus a few minutes) after the corresponding equinox or solstice in the previous year. Subtract one day in the case of a leap year. The reason for the addition of 5 hours and 49 minutes is because a tropical year (the mean time from one March equinox to the next) is 365 days, 5 hours, 48 minutes, and 45 seconds. The plus or minus a few minutes is a consequence of the Moon.
The timing of perihelion and aphelion is also conceptually simple. Add 6 hours and 9 minutes, then account for leap years, and then give a random toss of the dice to add or subtract a few days. The reason for adding 6 hours and 9 minutes (as opposed to 5 hours and 48 minutes) is because the sidereal year is slightly longer than is the tropical year. The reason for the apparent randomness is once again the Moon.
The Moon's orbit about the Earth is inclined by about 5 degrees with respect to the Earth's orbit about the Sun. That small inclination means that the Moon has but a small influence on the Sun's declination, which is what defines the equinoxes and solstices. On the other hand, that small inclination means the Moon has a significantly larger influence on the timing of Earth perihelion and aphelion.
A: If you just want to find the dates for various events, the the following link provides a list of sources for astronomical calculators of various types.
http://www.midnightkite.com/index.aspx?URL=Software
The US Naval Office also is a good source of this sort of astronomical information.
http://aa.usno.navy.mil/index.php
From the data you present in your graphic, I think it is fairly obvious that the motion of the earth around the sun is more complicated than you would expect, if modelled as a single body in the solar system, even over fairly short time scales.  I'm no expert in this, but I guess that the moon has a large part to play in the year to year variations in the various dates you show.
A: As David Hammen points out, there is nothing special about the points B and C. But if you insist, those are the points where the eccentric anomaly $E$ is $\pi/2$ and $3\pi/2$, respectively. For an idealized orbit, Kepler's equation gives
$$
\frac{2\pi}{T}(t-\tau) = E - e\sin E,
$$
where $\tau$ is the moment of periapsis (January 4) and $T=365.2596$ days is the anomalistic year. So
$$
(t-\tau)_\text{B} = \left(\frac{1}{4}-\frac{e}{2\pi}\right)T = 90.344\text{ days after periapsis}
$$
and
$$
(t-\tau)_\text{C} = \left(\frac{3}{4}+\frac{e}{2\pi}\right)T = 274.916\text{ days after periapsis}
$$
Of course, this is an idealized version. In reality Earth's orbit is subject to various perturbations.
