Definition of the interval
One possible definition given a strictly positive interval $[n,m]\subseteq\mathbb R^+$ could be:
$$
\prod_{x\in [n,m]}x:=\exp\left(\int_n^m\ln(x)\ dx\right)
=\frac{m^m\cdot n^{-n}} {\operatorname e^{m-n}}
$$
for an interval $[n,m]$ containing uncountably many elements. The countable and finite versions could then read
$$
\prod_{k=1}^{\infty} x_k:=\exp\left(\sum_{k=1}^{\infty}\ln(x_k)\right)
\quad\text{and}\quad
\prod_{k=1}^n x_k:=\exp\left(\sum_{k=1}^n\ln(x_k)\right)=x_1\cdot x_2\cdot ...\cdot x_n
$$
We could even extend this definition to
$$
\prod_{x\in [n,m]}f(x):=\exp\left(\int_n^m\ln(f(x))\ dx\right)
$$
One nice property is that with this defition we have
$$
\prod_{x\in[n,m]} x^a=\left(\prod_{x\in[n,m]} x\right)^a
$$
so it appears to follow some nice rules of powers of conventional finite products.
To define it without direct use of integration, my definition should be equivalent to the defining:
$$
S_k:=\prod_{x=0}^{2^k}\left(\frac{(2^k-x)n+xm}{2^k}\right)^{(m-n)/2^k}
$$
and recognize the product $\prod_{x\in [n,m]}x$ as the limit of those $S_k$'s as $k$ tends to infinity. So it is like multiplying together $2^k+1$ evenly spread out factors over the interval $[n,m]$, but adjusting the exponent of each factor to match the distance between the factors, namely $(m-n)/2^k$. These exponents tend to zero as the number of factors tends to infinity. It can then be shown that $S_k\to m^m\cdot n^{-n}/\operatorname e^{m-n}$.
So how does this relate to your suggested symmetrical expression? Well, you are considering the sequence of the form:
$$
T_k:=(S_k)^{2^k/(m-n)}=\prod_{x=0}^{2^k}\frac{(2^k-x)n+xm}{2^k}
$$
Now clearly, if $S_k\to a>1$ we will have $T_k\to\infty$ whereas for $S_k\to a\in[0,1)$ we must have $T_k\to0$. So the difficult cases are $S_k\to 1$ or $S_k\to a<0$. The latter I doubt we can make any sense of.
Resolving when $S_k$ tends to $1$
If $S_k$ tends to $1$ for some $n\in(0,1)$ and an appropriate matching $m>1$ we then know that
$$
\int_n^m\ln(x)\ dx=\lim_{k\to\infty}\ln(S_k)=\ln(1)=0
$$
Considering the graph of $\ln(x)$ it can be shown that
$$
\left[\ln m -q_k\cdot(\ln m -\ln n)\right]\cdot\Delta x\leq\ln(S_k)\leq\left[\ln n +(1-q_k)\cdot(\ln m -\ln n)\right]\cdot\Delta x
$$
where $0.5\leq q_k\leq 1$ is a sequence tending to $0.5$, and $\Delta x$ is short hand for the distance $(m-n)/2^k$. Since $T_k$ is equal to $S_k$ except for raising to the reciprocal of $\Delta x$ we get $\ln(T_k)=\ln(S_k)/\Delta x$ and therefore
$$
\left[\ln m-q_k\cdot(\ln m -\ln n)\right]\leq\ln(T_k)\leq\left[\ln n +(1-q_k)\cdot(\ln m -\ln n)\right]
$$
As $k$ tends to infinity both bounds tend to $(\ln m+\ln n )/2$ showing us that
$$
\lim_{k\to\infty} T_k=\exp\left(\frac{\ln m+\ln n}2\right)=\sqrt{n\cdot m}
$$
Now we should able to do the following:
Resolving the computation method
Given $n\in(0,1)$ one can solve
$$
\int_n^m\ln(x)\ dx=m(\ln m-1)-n(\ln n-1)=0
$$
for $m>1$ in order to find the corresponding $m$ so that $T_k$ converges. One way to do this is to use the Newton-Raphson method on the function
$$
f(x)=x(\ln x-1)-n(\ln n-1)
$$
with initial guess $x_0=2$. Then $m=\lim_{k\to\infty}x_k$ where the $x_k$'s are defined recursively as
$$
x_{k+1}:=x_k-\frac{x_k(\ln x_k-1)-n(\ln n-1)}{\ln x_k}
$$
It turns out that in fact $1<m<\operatorname e\approx 2.7182818$. For $n=0.5$ it takes only a few iterations before one has
$$
m\approx x_5=1.603016489916967074791...
$$
and it can be verified that the digits listed above do not change for future iterations so we already have $m$ to a very high precision. So we have
$$
\lim_{k\to\infty}\prod ^{2^k} _{x=0} \frac{(2^k - x)0.5+x\cdot 1.603016489916967074791...}{2^k}\\=\sqrt{0.5\cdot 1.603016489916967074791...}\approx 0.8952699285458456285
$$
But this is a very unstable result anyway! My earlier computations showed that if $m$ is either the slightest bit larger than the actual solution to $f(x)=0$ the product tends to infinity, or if it is the slightest bit less then it tends to zero.
Some general remarks to conclude
The infinite symmetrical product you defined is very unstable in more than one respect. If we change the definition even slightly we may get an entirely different result:
$$
\prod ^{2^k} _{x=1} \frac{(2^k - x)n+xm}{2^k}
$$
removes the first factor $n$ whereas
$$
\prod ^{2^k-1} _{x=0} \frac{(2^k - x)n+xm}{2^k}
$$
removes the last factor $m$. Whereas these would both lead to the same values of $S_k$ removing only a negligible contribution at that level, they affect the value of $T_k$ by a non-negligible factor. Also if we distributed the factors in the interval $[n,m]$ slightly different the product $T_k$ could change a lot whereas $S_k$ would not. So overall $S_k$ is a much more stable value. And $S_k$ tends to represent an actually uncountable product, whereas $T_k$ tends to something I would characterize as product of a countable subset of the factors in question. This might be another reason it is so unstable - there are infinitely many other ways we could have defined $T_k$ that would yield totally different results whereas all definitions of $S_k$ by partitioning $[n,m]$ into subintervals that decrease toward zero width as $k$ tends to infinity would all point to the same limit value of $S_k$. This in a sense addresses your question 1 as an infinite product of the kind you are suggesting would or would not tend to zero depending on the distribution you use to select factors from $[n,m]$.
The notion of "countably infinite" is well named. Another word you can use is "enumerable," which is even more descriptive in my opinion.
I understand your intuition on the subject and I see where you're coming from. Let me try to give you some insight into the agreements about infinity that have been reached over the years by the mathematicians who've worked on this problem. (This is what I wish someone had explained to me.)
"Countably infinite" just means that you can define a sequence (an order) in which the elements can be listed. (Such that every element is listed exactly once.)
The natural numbers are the most naturally "countable"—they're even called "counting numbers"—because they are the most basic, natural sequence. The word "sequence" itself is wrapped up in what we mean by natural numbers, which is just one thing following another, and the next one coming after that, and the next one after that, and so on in sequence.
But the integers are countable as well. In other words, they are enumerable. You can specify an order which lists every integer exactly once and doesn't miss any of them. (Actually it doesn't matter, for the meaning of "countable," if a particular element gets listed more than once, because you could always just skip it any time it appears after the first time.)
Here is an example of how you can enumerate (count, list out) every single integer without missing any:
$0, 1, -1, 2, -2, 3, -3...$
The rational numbers are also countable, again because you can define a sequence which lists every rational number. The fact that they are listed means (at the same time) that they are listed in a sequence, which means that you can assign a counting number to each one.
The real numbers are not countable. This is because it is impossible to define a list or method or sequence that will list every single real number. It's not just difficult; it's actually impossible. See "Cantor's diagonal argument."
This will hopefully give you a solid starting point to understanding anything else about infinite sets which you care to examine. :)
Best Answer
In algebraic sense, it is just an undefined quantity. You have no way to assign a value to $\tan (\pi/2)$ so that its value is compatible with usual algebraic rules together with the trigonometric identities. Thus we have to give up either the algebraic rules or the definedness of $\tan(\pi/2)$. In many cases, we just leave it undefined so that our familiar algebras remain survived.
But in other situation, where some geometric aspects of the tangent function have to be considered, we can put those rules aside and define the value of $\tan(\pi/2)$. For example, where we are motivated by the continuity of the tangent function on $(-\pi/2, \pi/2)$, we may let $\tan(\pm\pi/2) = \pm\infty$ so that it is extended to a 1-1 well-behaving mapping between $[-\pi/2, \pi/2]$ and the extended real line $[-\infty, \infty]$. In another example, anyone who is familiar with complex analysis will find that the tangent function is a holomorphic function from the complex plane $\Bbb{C}$ to the Riemann sphere $\hat{\Bbb{C}}=\Bbb{C}\cup\{\infty\}$ by letting $\tan(\frac{1}{2}+n)\pi = \infty$ the point at infinity.
In conclusion, there is no general rule for assigning a value to $\tan(\pi/2)$, and if needed, it depends on which property you are considering.