Why can't I calculate 1.5 standard deviations using basic math? 
I don't understand why I can't simply add 1.5 standard deviations to get the answer.
If 1 standard deviation is 10kg and the mean is 400kg, then 415kg is 1.5 standard deviations.
So I calculated it like this: .3413 + ((.4772-.3413)/2) = 0.40925
This equation takes one half of the difference between two standard deviations and one standard deviation, then adds it to the first standard deviation.
Why does this not work? Why do I have to use the table provided?
 A: Just to provide a different illustration on the same topic...
In your initial calculation you would be treating the normal curve as a uniform distribution, in which case your initial approach would be the correct mathematical calculation for the double hatched rectangle in the plot below (with different actual values), simply because you'd be able to express the area as a simple linear dependency of the $x$ axis distance:
$A_{1.5\,SD} =\large\frac{A_{2\,SD} - A_{1\,SD}}{2} = \small height * \large\frac{X_{2\,SD} - X_{1\,SD}}{2} $
But you want to calculate the diagonally hatched area under the curve of the Gaussian distribution, which as stated before wouldn't keep a linear relationship with the distance along the $x$ axis even if the distribution was triangular:

7
A: The formula for the Gaussian distribution is:

Where sigma = std deviation and mu = mean
(stolen from wikipedia)
When you are asking for the area, you are integrating this function over the range specified.  This integral does not have a "closed form" solution: there is no way to come up with an expression using "normal" math functions like factorial, multiplication, exponentiation, roots, etc. that equals that integral.
It's just like logarithms or trigonometric functions: you can't produce a closed form equation for them using other algebraic functions (you can use infinite series, but that's not "closed").  So you use a table (if you are feeling retro, or a calculator, which simply uses a table for you behind the scenes embedded in its processor as a starting point) when you need to actually calculate it.
In fact, the parallel with logarithms is quite apt: one can also define a logarithm by an integral, namely ln(x) = integral of (1/x) from 0 to x.
A: The reason that we cannot (linearly) interpolate between 0.3413 and 0.4772 is because the pdf of the Normal distribution is not uniform (flat at a single value).
Consider this more simple example, where we can use geometry to find the areas. 

The total area of the plot is 1 (it's a square cut diagonally, with the two pieces rearranged to be a triangle). Using Base*Height/2 we can find that the area of region A is 0.5, and the total area of regions B and C is also 0.5.
But the areas of B and C are not equal. The area of region C is 0.5*0.5/2 = 0.125, and therefore the area of region B is 0.375. So even though regions B and C are equally wide along the x-axis, since the height is not constant, they have different areas.
The Normal distribution that you are dealing with in your exercise is similar, but with a more complicated function for the height instead of a simple triangle. Because of this, the area between two values can't be solved as simply - hence the use of Z-scores and a table to find probabilities.
A: Geometrically, .4772 - .3413, represents the area under the graph between 1 standard deviation and 2 standard deviations.  If you split this region half way across horizontally, the part to the left of the split will be the area between 1 and 1.5 standard deviations, as you want.  Fine so far.
However when you take (.4772 - .3413) / 2 you're getting half the area, but not necessarily what you were looking for, which is however much area was half way across horizontally. With this graph, that left part of the split isn't half of the area - the line is sloping downward (going from the top left to the bottom right) so there's more space in the left part than the right part.  If this graph was a straight horizontal line, then the area you were splitting would be a rectangle, and half the area really would be half way across.
