Suppose I placed a block on a table and lent on the top of it. There would be a downwards force on the block due to its weight, and a further downwards force on the block due to my arm. But the block wouldn't be moving --- it wouldn't be accelerating in the slightest. And so, by Newton's second law, there must be no force acting on it! What this means is that the table must be exerting a force equal in magnitude to the sum of both the weight of the block and the force from my arm, but upwards, so as to cancel everything out.
Hopefully this is familiar to you. The point of all this is that normal forces (and more generally, constraint forces) vary in just such a way as to maintain a certain condition --- in this case, it's the condition that the block doesn't sink into the table. You might be used to $N = mg$ or something similar, but that doesn't have to be so. Indeed, if I attached a string to the top of the block and pulled slightly on the end, the normal force would be less than $mg$. This is because we already have an upwards force acting on the block --- that due to the string --- and so the table doesn't have 'as much ground to make up' when it comes to making sure all the forces are cancelled.
With this said, let's approach the two systems you're looking at. You asked: "why is the normal force not $F_g \cos \theta$ like it was in the last problem?". I could ask you: why should it be the same? The two systems are rather different: one involves a stationary block, the other a ball undergoing circular motion, and we've seen that normal forces can vary from circumstance to circumstance. A priori there's no reason to assume that the normal forces should have the same form. But let's consider specifically why they don't.
In the first case, the block isn't moving and so (since its acceleration is therefore zero) the net force is zero. Or at the very least (suppose that in actual fact the block was sliding along the plane) we can say that the block isn't moving in the direction perpendicular to the plane, and so the forces in that direction add to zero. The component of the force of gravity in this direction is $F_g \cos \theta$, whereas the component of the normal force in this direction (which is just the total normal force) is $F_N$. These forces must cancel, and so
$$ F_N - F_g \cos \theta =0\,.$$
In the second case, the block is moving and in fact it is accelerating. It's accelerating because it's constantly changing direction. It might seem counter-intuitive, but the direction of this acceleration is actually towards the centre of the circle. What this means is that the components of the forces in the horizontal direction won't cancel; they would only cancel if our acceleration were zero in this direction. As I've said, and as the diagram you gave suggests: there is acceleration in this direction.
However, there isn't any motion in the vertical direction. This is true essentially by assertion --- we're considering a problem in which a ball rolls around a banked curve at constant height. If there's no motion in the vertical direction, the components of the forces in the vertical direction must sum to zero. We see from the diagram that the only forces acting are the normal force and the weight of the ball. So, resolving the vectors correctly, we find that we must have
$$F_N \cos \theta - F_g = 0$$
if indeed the vertical forces are to sum to zero, as we require. The other equation --- that for components in the horizontal direction, will look like this
$$ F_N \sin \theta = ma \,.$$
So to check we understand: if we were to resolve forces perpendicular to plane in this second example, we couldn't conclude that the forces we obtained summed to zero. The reason? Well, the ball has an acceleration that points in the horizontal direction, and so the component of its acceleration along the direction perpendicular to the plane is non-zero. As such, the net force in this direction is non-zero. It may not look like the ball has a component of acceleration perpendicular to the plane --- that would seem to suggest that the ball was trying to lift off! But, despite how it might seem, the fact of the matter is that the ball has an acceleration that points horizontally, and so there is a component of this acceleration along the direction perpendicular to the plane.
Hope this helps!