You can start a summation or a product anywhere, not just at $1$. There is no law mandating that they must start at $1$.
In this case, if you were to allow $i=1$, then the first factor of the product would be $\left(1 - \frac{1}{1^2}\right) = 0$ and so the entire product would be zero, which would make everything rather silly. So you start at $2$ instead (you could instead reindex and consider $\prod_{j=1}^{n-1}\left(1 - \frac{1}{(j+1)^2}\right)$ but that just makes the expression look more complicated; better to change the indices instead).
Why do you have $n\geq 2$? Because if $n\lt 2$, then the product has no factors (since no value of $i$ can be at least $2$ and also less than or equal to some $n\lt 2$). The empty product is equal to $1$ by definition, but most people have trouble with that assertion, so why introduce it? Better to stick to $n\geq 2$ so that it is clear there are always factors.
Yes: you start by plugging in $i=2$, then $i=3$, then $i=4$, and so on until you get to $i=n$. For example, with $n=4$, you have \begin{align*} \prod_{i=2}^4\left(1 - \frac{1}{i^2}\right) &= \left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{3^2}\right)\left(1 - \frac{1}{4^2}\right)\\ &= \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right)\\ &= \left(\frac{3}{4}\right)\left(\frac{8}{9}\right)\left(\frac{15}{16}\right)\\ &= \frac{360}{576} = \frac{5}{8}. \end{align*} As to finding a formula, well, you might try a few values and see if you spot a pattern; then you can try using induction to prove it.
Added. Okay, you say you are stuck, you don't see a pattern. This is just
Consider the factors you are multiplying: $1 - \frac{1}{i^2}$. Writing them as a single fraction, you have $1 - \frac{1}{i^2} = \frac{i^2 - 1}{i^2} = \frac{(i-1)(i+1)}{i^2}.$
So, what are you doing at each step? Let's write them out explicitly: $\left(\frac{(1)(3)}{2^2}\right)\left(\frac{(2)(4)}{3^2}\right)\left(\frac{(3)(5)}{4^2}\right)\cdots\left(\frac{(n-1)(n+2)}{n^2}\right).$ The denominator is easy: you can just rearrange it, throw in a couple of extra factors of $1$ (who cares if you multiply by $1$? Doesn't do anything) and write: \begin{align*} &(2)(2)(3)(3)(4)(4)\cdots(n)(n)\\ &=(2)(3)(4)\cdots(n)(2)(3)(4)\cdots(n)= \Bigl((1)(2)(3)(4)\cdots(n)\Bigr)\Bigl((1)(2)(3)(4)\cdots(n)\Bigr). \end{align*} Now, what do you call the product of all the positive integers from $1$ through $n$?
Great! We now have a formula for the denominator.
What about the numerator? Again, reorder them putting first all the first factors, and next all the second factors; we have: \begin{align*} &(1)(3)(2)(4)(3)(5)(4)(6)\cdots (n-1)(n+1)\\ &= \Bigl( (1)(2)(3)(4)\cdots(n-1)\Bigr)\Bigl((3)(4)(5)\cdots(n+1)\Bigr). \end{align*} Hmmm... Looks like another set of those "products of all positive integers from $1$ up to something
." Now, granted, the second factor isn't that, but just throw in a $(1)(2)$, and then divide by $2$ to keep it honest: $\Bigl((1)(2)(3)(4)\cdots(n-1)\Bigr)\Bigl((1)(2)(3)(4)(5)\cdots(n+1)\Bigr)\left(\frac{1}{2}\right).$ Now, express it using the same kind of symbol as the denominator.
Now, step back and look at what you get. Is there some simplification/cancellation that you can make to get something simple? What?