The first identity is provable simply by multiplying out the brackets:
(1-x)(1+x+x^2+x^3) = 1+x+x^2+x^3-x-x^2-x^3-x^4 = 1-x^4
We see that all the middle terms in x cancel out, so in general,
(1-x)(1+x+x^2...+x^n) = (1-x^(n+1))
As n->inf, as long as abs(x)<1, x^n -> 0. Therefore, as n->inf, the expression tends to 1.
The second identity is provable because it is a repeated application of the difference of two squares:
(1-x^8) = (1-x^4)(1+x^4) = (1-x^2)(1+x^2)(1+x^4) = (1-x)(1+x)(1+x^2)(1+x^4)
If we were to repeat this process by continuing multiplication by (1+x^(2^n)), we would end up with:
(1-x)(1+x)(1+x^2)...(1+x^(2^n)) = (1-x^2)(1+x^2)...(1+x^(2^n))... = (1-x^(2^(n+1))).
For the same reason as the previous example, as n->inf, this tends to 1.
The third identity is provable by multiplying both sides by 16sin(x/16). This gives us 2sin(x/16)cos(x/16)*8cos(x/8)cos(x/4)cos(x/2) = sin(2x/16)*8cos(x/8)cos(x/4)cos(x/2) = 2sin(x/8)cos(x/8)*4cos(x/4)cos(x/2)... which by repeating the method of collapsing the double-angle formula for sin(2theta) comes to sin(x).
In general, then, this can be repeated so product(cos x/2^k, k=1 to n) = sin(x)/(2^n sin(x/2^n)), except when sin(x/2^n) is equal to 0, which is when x = 2^n pi or a multiple of this.
As n->inf, the left hand side gets multiplied by cos(closer to 0), ie. by (value closer to 1). Therefore it looks likely that there is a limit. The only way I could find to discover this limit was applying L'Hopital's rule on the RHS to give cos(x)/cos(x/2^n); as n->inf, the quantity on the bottom of the fraction becomes 1, so the limit is cos(x). This is valid for all x, I think.