Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$
We have \begin{align*} p_{k} &= \prod_{n = 2}^{k} \left( 1 - \frac{1}{n^{2}} \right) = \prod_{n=2}^{k} \frac{(n-1)(n+1)}{n^{2}} \\ & = \frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{2\cdot 4}{3 \cdot 3} \cdot \frac{3\cdot 5}{4 \cdot 4} \cdot \cdots \cdot \frac{(k-2)\cdot k}{(k-1)\cdot(k-1)} \cdot \frac{(k-1)(k+1)}{k\cdot k}\\ & = \frac{1}{2}\left(1 + \frac{1}{k}\right) \end{align*} because all but the first and last numerators and denominators cancel. Therefore \[ \prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^{2}} \right) = \lim_{k\to\infty} p_{k} = \frac{1}{2}. \]
To put this into a little context: Euler has shown that \[ \sin{(\pi z)} = \pi z \prod_{n=1}^{\infty} \left( 1 - \frac{z^{2}}{n^{2}} \right) \] for all $z \in \mathbb{C}$.
We would like to plug in $z = 1$ on both sides. This doesn't work because we get $0 = 0$. But a simple trick yields what we want: \[ \pi \prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^{2}} \right) = \lim_{z \to 1} \frac{\sin{(\pi z)}}{(1- z^2)} = \lim_{z \to 1} \frac{\pi \cos{(\pi z)}}{-2z} = \frac{\pi}{2} \] and cancelling on both sides with $\pi$ gives $\frac{1}{2}$ for the infinite product.
There is a lot of fun that you can have with Euler's product. For instance $z = \frac{1}{2}$ yields the Wallis product formula \[ \frac{\pi}{2} = \frac{2 \cdot 2}{1 \cdot 3} \cdot \frac{4 \cdot 4}{3 \cdot 5} \cdot \frac{6 \cdot 6}{5 \cdot 7} \cdot \cdots = \prod_{n=1}^{\infty} \frac{2n}{2n -1}\frac{2n}{2n +1} \] and $z = i$ yields \[ \frac{\sin{(i\pi)}}{i \pi} = \frac{e^{\pi} - e^{-\pi}}{2\pi} = \prod_{n = 1}^{\infty} \left( 1 + \frac{1}{n^{2}} \right). \] A thorough treatment of these and many other topics involving infinite products can be found in chapters 1 and 2 of Remmert, Classical topics in complex function theory, Springer GTM 172, 1998. The title of the German original is simply Funktionentheorie 2.
The product can never be 1. Recall that the product is defined as the limit of the partial products $$ \prod_{n\le k}\left(1-\frac1{f(n)}\right)=A_k. $$ Now, if $f(n)=1$ then the product is 0. I assume you know how to handle "divergence to 0"? To simplify, let's assume $f(n)\ne 1$ for all $n$.
Then we have that $A_k>A_{k+1}$ for all $k$, since $1-\epsilon<1$ for any positive $\epsilon$ (say, $\epsilon=1/f(k+1)$). Since $A_1=1-1/f(1)<1$, our sequences will converge to a positive number smaller than 1, or diverge to 0, but it most certainly cannot converge to 1.
Here is a hint to evaluate $$\prod_{n=2}^\infty\left(1-\frac1{n^2}\right) :$$ Note that this is a telescoping product, since $1-1/n^2=(n-1)(n+1)/n^2$. Now play with the first few terms to see the emerging pattern.
Your friend is right: the partial products are clearly $\le 3/4$, and since the infinite product converges (by the test you showed), its value must also be $\le 3/4$. To find the actual value, note that $1 - 1/n^2 = (n-1)(n+1)/n^2$, and so the infinite product is $$ \prod_{n=2}^{\infty}\left(1 - \frac{1}{n^2}\right) = \frac{1}{2}\frac{3}{2}\cdot\frac{2}{3}\frac{4}{3}\cdot\frac{3}{4}\frac{5}{4}\cdot\frac{4}{5}\frac{6}{5}\cdot ... $$ All terms cancel in pairs except the first; so the value of the product is $1/2$.