I'll try to convince you that they are geometrically quite obviously different, but when it comes to naming them, the begin to look alike :)
Geometrically you probably already have a good picture of what a point is: it's just the primitive notion of a point you have in geometry. That is, a single dimensionless location in space.
A vector should be thought of as having two qualities: a ray that has direction and magnitude. In basic vector algebra in $\Bbb R^n$, we learn that such a ray can slide all around $\Bbb R^n$, and as long as you aren't changing the direction or the length of the ray, then it is still the same vector.
Now when it comes to naming these two things, they start to look alike! With Cartesian coordinates, points in $\Bbb R^n$ are labeled by their projections to the axes, and that creates a list of real numbers. Similarly, when we go about naming vectors, we have this convention of sliding the vector so that it is being emitted from the origin, and then we check to see what point is on its arrowhead. The vector is named after this point.
So in both cases, a similar list of real numbers is used to identify the object. Since this is the case, it's common to just start referring to any ordered $n$-tuple of things from a field (like $\Bbb R$) as a "vector," even if we aren't thinking of it as a ray in that application.
One example is that of vector fields. Since these are functions of position, the inputs they take are points of $\Bbb R^n$ (which look like an ordered $n$-tuple). The outputs are vectors (which again look like an ordered $n$-tuple), but we are interpreting these as the vectors they represent, slid over from the origin to the point we're at.
You can, of course, really have vector inputs! For instance, the length of a vector in $\Bbb R^n$ creates a function from vectors into $\Bbb R$. Of course, the same function could be reinterpreted as the distance-to-zero function on points of $\Bbb R^n$.
So, the difference is all in how you are interpreting that particular list of numbers.
For #1 in your post, you are probably thinking of it as the line segment between points $x$ and $x'$. The addition that's going on is vector addition though. Drawing the vectors that $x$ and $x'$ represent, you see you have two vectors extending from the orign to these two points. For any two vectors $v,w$, $v-w$ yields the vector which fits between the two tips of $w$ and $v$, and points to the tip of $v$. So, you can see that $x-x'$ has the point $x$ on its tip.
What does the $t$ contribute? If you multiply out $xt+(1-t)x'=x'+t(x-x')$, you can see that the vector $x-x'$ is being scaled by $t$ to something shorter, and then is being concatenated onto the tip of $x'$. The tip of this arrow gives another point on the segment. Ranging over all $t$ between 0 and 1, you get vectors pointing to all points on that segment.
What determines the radius and center of the resulting sphere?
Because the condition is defined in terms of points $\mathbf a$ and $\mathbf b$ only, the points satisfying it have to be symmetrically distributed about the line through $\mathbf a$ and $\mathbf b$. Which means that the center of the sphere must lie on this line. There are two points on this line that satisfy the condition themselves: one is two thirds of the way from $\mathbf a$ to $\mathbf b$, and the other is on the opposite side of $\mathbf b$ from $\mathbf a$, at the same distance away. These two points have to be the opposite sides of a diameter, so the center is halfway between them, and the radius is half the distance between them, which is $2/3$ the distance from $\mathbf a$ to $\mathbf b$.
If I ask instead for the set of all points which are equidistant from A and B, I get a plane. If I ask for the same in $\mathbb R^2$, I get a line. If I ask for all points equidistant from just one point in $\mathbb R^3$, though, I get a sphere. Is it true in general that the set of all points in $\mathbb R^n$ equidistant from k selected points is a subspace of dimension $n−k$?
Look carefully again at your examples. Both examples in 3-space are sets of dimension 2, even though one is a distance from two points and the other from one point. And for the examples in 2-space, both are of dimension 1, even though one is a distance from two points and the other from one point.
There are some rules here, but not what you gave. Generally if you have a variable defined on a space and then ask for the set where that variable has a specific value, you can expect that set to be 1 dimension less than the space (we say "of codimension 1") if you have two independent variables and specify values for both of them at the same time, you get something of codimension 2, etc.
If I don't ask for equidistant points but instead ask for something like the above, i.e some distances are proportional to some other distances, what is the analogous result?
The ratio is a variable that you are proscribing a value for, so it still defines a set of codimension 1.
Best Answer
You don't add locations, you add displacements.
Even as a kid you would not be taught scalar addition as adding points on the number line, but adding the distance moved between points. Start at the origin ($0$), move distance $x$, then move distance $y$ from there and the point you arrive at has the value of the sum of $x$ and $y$.
That's basically adding 1-dimensional displacement vectors.
Are you sure that you understand the distinction between points and vectors?