Given only the few facts you are allowed to use, there is some ambiguity concerning what the value of $\cos(\pi/3)$ might be.
As far as I understand, you are supposed to consider two functions $f$ and $g$ such that for all $x$ and $y,$
\begin{align}
f(x)^2+g(x)^2&=1, \\
f(x+y)&=f(x)g(y)+g(x)f(y), \\
f(x-y)&=f(x)g(y)-g(x)f(y), \\
g(x+y)&=g(x)g(y)-f(x)f(y), \\
g(x-y)&=g(x)g(y)+f(x)f(y), \\
f(x+2\pi)&=f(x),\\
g(x+2\pi)&=g(x).\\
\end{align}
The idea (I suppose) is that if we define $f$ and $g$ this way, then $f$ is the sine function and $g$ is the cosine function.
Now if you apply the difference formula to
$g(x-x),$ you can quickly find (by using the Pythagorean identity) that $g(0)=1.$
Again applying Pythagoras you then have $f(0)=0.$
But notice what happens if we set $f(x)=0$ and $g(x)=1$ for all $x,$
that is, if $f$ and $g$ both are constant functions.
All of the equations you were given are satisfied.
So technically I do not think the facts you are allowed to use are enough to distinguish the sine and cosine from constant functions, let alone find the value of $\cos(\pi/3).$
The reason I said the question is ambiguous is because although any constant function is periodic with period $2\pi,$ that is not its minimum period.
If you are given that the minimum period of each of the functions $f$ and $g$ is $2\pi,$ by applying the sum formula to $g(\pi+\pi)$ and then the Pythagorean identity to eliminate $f,$
you can show that $(g(\pi))^2=1.$
Furthermore, if $g(\pi)=1$ then you can show that $g(\pi+x)=g(x),$ so $g$ would have period $\pi.$
Since that is less than the minimum period of $g,$ we rule out $g(\pi)=1.$ So we must have $g(\pi)=-1$ instead.
With that fact established, you can proceed as in some of the other answers.
Even with the “minimum period” stipulation, it’s a good thing you were asked for a cosine value, because the facts you were given are still not sufficient to tell whether $f(x)=\sin(x)$ or $f(x)=-\sin(x)$.