Your new teacher is wrong. $\sqrt{\cdot}$ is the principal square root operator. That means it returns only the principal root -- the positive one. $\sqrt{64}=8$. It does NOT equal $-8$.
On the other hand, the equation $64=x^2$ DOES have $2$ solutions: $x=8$ or $x=-8$. Thus both $8$ and $-8$ are square roots of $64$.
Let's see what happens when we take the principal square root of both sides of this equation: $$\begin{align}64 &= x^2 \\ \implies \sqrt{64} &= \sqrt{x^2} \\ \implies 8 &= |x| \\ \implies x&=8 \text{ or } x=-8\end{align}$$
Thus the fact that the principal square root operation throws out the negative root isn't much of a problem as the math still works out correctly.
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 $[n,m]$.
Best Answer
This is an interesting question, but thanks to the properties of exponentiation, it can be easily solved as follows: $$\prod_{k\ge1}\sqrt[k]{n}=\prod_{k\ge1}n^{1/k}=n^{\sum_{k\ge1}\frac{1}{k}},$$ and as the exponent diverges (Harmonic series) the whole expression diverges, as long as $n>1$.
EDIT: the first equality is definition. For the second equality we need to argue why one can pass to the limit.
In other words: for any fixed $N$ the equality $$\prod_{k=1}^N n^{1/k}=n^{\sum_{k=1}^N \frac{1}{k}},$$ is valid, but why does it hold that $$\lim_{N\rightarrow \infty}n^{\sum_{k=1}^N \frac{1}{k}}=n^{\lim_{N\rightarrow\infty}\sum_{k=1}^N \frac{1}{k}}\quad ?$$ In this particular case it is a consequence of the fact that the map $x\mapsto n^x$ diverges to $+\infty$ as $x\rightarrow +\infty$ and by abuse of notation we write $n^{+\infty}=+\infty$
If conversely the exponent would converge to a finite limit, as for example in the case mentioned in the comment of IanF1, then this would be a simple consequence of the fact that the map $x\mapsto n^x$ is continuous.