Are vector-valued functions like functions of a single variable with
domain of real values and range of vectors in an n-dimensional space?
Yes, but a vector-valued function can have any number of variables, e.g., ${\bf v}(r, s, t, w)$.
Does the vector-valued function involve a formula for a position
vector which traces out a curve when the range of input values varies?
Yes, but again, there may be several variables involved. For a continuous vector-valued function the "tip of the vector" in general sweeps out a line, surface, volume, etc., depending upon the number of variables.
Do we use parameterisation's to describe curves which cannot be
expressed as functions like y=f(x) and x=g(y)? How are vector-valued
functions and parameterisation's related? What are parametric equation and are they similar to vector valued functions in any way?
One can parameterize a vector-valued function as ${\bf v} = (v_1, v_2, \ldots, v_k) = (f_1(t), f_2(t), \ldots, f_k(t))$.
You are right that $\frac xx$, or rather the function conventionally meant by that term, is continuous. But I think there is a conceptual issue at play here, and a common one. Namely, that the functional equation
$$f(x)=\textrm{some expression in }x$$
already determines the function. This is not true. When building a function, we specify things in this order: domain and codomain first, functional equation second. That is, we choose the domain and codomain of the function, and then we are bound to give a functional equation which is actually defined for all elements of the domain, and returns only elements of the codomain. In mathematical notation you'll see it like $f:X\to Y,~f(x)=\dots$, where $X,Y$ are domain and codomain, and they are specified first. Now this said, $f:\mathbb R\to\mathbb R,~f(x)=\frac xx$ is not a well-defined function, since the functional equation isn't meaningful for all elements of the domain. However, $f:\mathbb R\backslash\{0\}\to\mathbb R,~f(x)=\frac xx$ is well-defined, since the functional equation is meaningful for all elements of the previously specified domain, and it returns only elements of the previously specified codomain. So to be sensible, when we write "the function $\frac xx$", we usually mean that as shorthand for the latter definition of a function. But technically speaking, just the term $\frac xx$ does not define a function, since we haven't specified domain or codomain, which we must, unless we want our readers to guess (which isn't always unfeasible, to be fair).
Now with all this about the definition of functions out of the way, continuity can be treated more clearly: a function $f:X\to Y$ is continuous at $x\in X$ if [insert your favorite definition of continuity at a point]. A function $f:X\to Y$ is continuous if it is continuous at all $x\in X$.
Notice that this definition only talks about points inside the domain $X$, at which the function must be defined in order to be a function. Whatever lies outside of the domain is of no concern for the definition of continuity.
As a closing remark, the often cited example of the discontinuous function $\frac 1x$ is just bad, because $f:\mathbb R\to\mathbb R,~f(x)=\frac 1x$ is not a function, while $f:\mathbb R\backslash\{0\}\to\mathbb R,~f(x)=\frac 1x$ is a perfectly continuous function. There just isn't any sensible (meaning non-contrieved) way to interpret $\frac1x$ as something that is both a function and discontinuous.
Best Answer
A fundamental difference between differentiation and integration is that differentiation is performed at specific points, while integration is done over a region. So while you are correct that a function cannot be differentiated at "sharp points" (more formally, points where the function is not differentiable), there is nothing to stop you from integrating over sharp points (so long as your function is integrable on whatever domain you're integrating over).
Though harder to actually compute, integration is, in a sense, nicer than differentiation. A differentiable function is necessarily continuous, but this is not the case for integrable functions - in fact, any continuous function is integrable. So even if you do have "sharp points", so long as your function is continuous (and even if it's not in pretty much every case you'll ever encounter in high school), you can integrate over any points at which your function isn't differentiable.
Practically, if your domain of integration includes the point $x=1$, then that just means you'll need to split up your integral. For example: $$\int_0^2|1-x|\,dx=\int_0^1(1-x)\,dx+\int_1^2(x-1)\,dx.$$