The coordinate system, USA Contiguous Albers equal area conic, by default is in meters, so any calculations you do should be in meters or square meters. Depending on how large the polygons are, the values could be very large. Did you want to have the values reported in kilometers instead? The Calculate Geometry tool lets you choose a different unit value.
In the ArcGIS for Desktop 10.1 help, the help topic is in Geodata, Data types, Tables, Calculating field values, Calculating area, length, and other geometric properties.
There's a small chance that someone modified the USA Contiguous Albers definition and changed the units. You can check this by looking at the coordinate system of the layer in ArcMap or ArcCatalog.
If I understood correctly, by "real distance" you mean the distance along an actual 3D straight line connecting your two points. The term "geodesic" refers to a straight line on the curved ellipsoid surface.
Using the pythagorean theorem with the geodesic length and the height difference would not work, because the geodesic line is curved, it is not a square triangle with straight segments.
To calculate a 3D straight distance from lat/lon/ht, you will first need to convert your points to 3D cartesian coordinates (see formulas below). For modern reference systems like WGS84/ITRF2014, the reference frame is called an ECEF (Earth-Centered, Earth-Fixed). Then you will be able to calculate the distance with the square root method in 3D and get the correct answer. However keep in mind that a straight line in 3D, over long distances, can pass through the Earth, so this calculation is mainly useful for 3D "line of sight" types of problems for generally short distances.
Also, make sure you are using ellipsoidal heights, not altitudes above sea level or geoid level, since this could slightly alter the distances. If you need to convert from geoidal (mean sea level) to ellipsoidal, this tool can be useful.
Let's write the lat, lon, ht of a point as phi, lambda, h. You can find the X, Y, Z coordinates of the point like this :
X = (N(phi)+h) * cos(phi) * cos(lambda)
Y = (N(phi)+h) * cos(phi) * sin(lambda)
Z = ((b^2/a^2) * N(phi) + h) * sin(phi)
Where :
N(phi) = a / sqrt(1-e^2*sin^2 phi)
a = semi-major axis (6,378,137 m)
b = semi-minor axis (6,356,752.314245 m)
e^2 = 1 - (b^2 / a^2)
Then with pythagorean in 3D, we find the distance d:
d = sqrt{(X_2-X_1)^2 + (Y_2-Y_1)^2 + (Z_2-Z_1)^2}
Best Answer
I would not recommend trying to recover a latitude and longitude in order to "apply the distance formula for coordinates by hand to get the distances in miles" because the length of a degree of longitude varies depending on its latitude i.e. 0 m at poles, and about 110 km (= 70 miles) at around 30 degrees N or S.
I think you will be far better to get your distances in meters direct from using Near with your Albers coordinate system and then multiply them by 39.37 to get inches, and divide that by 63,360 to get miles.