If you are solving a multiple choice test like GRE you really need fast intuitive, but certain, thinking. I tried to put myself in this rushed set of mind when I read your question and thought this way: think of $x^{12}$ as something like $x^2$ but growing faster, think of $2^x$ as $e^x$ similarly, sketch both functions.
It is immediate to see an intersection point for $x<0$ and another for $0, for some positive $b$ since the exponential grows slower for small $x$ for a while, as the sketched graph suggests. So the answer is at least $2$. In fact it is $3$ because after the second intersection point you clearly see the graph of $x^{12}$ over $2^x$ but you should notice that $a^x\gg x^n$ at $+\infty$, and therefore the exponential must take over and have a third intersection point at a really big value of positive $x$. Once this happens the exponential function is growing so fast that a potential function cannot catch up so there are no further intersections.
(To quickly see that $a^x\gg x^n$ at $+\infty$ just calculate $\lim_{x\to\infty}\frac{a^x}{x^n}=+\infty$ using L'Hopital's rule or Taylor expanding the numerator whose terms are of the form $\log^m(a)a^m/m!$).
More rigorously, maybe you can find a way to study the signs of $g(x)=x^{12}-2^x$ using derivatives and monotony. There are 4 intervals giving signs + - + - resulting in 3 points of intersection by the intermediate value theorem. These intervals are straightforwardly seen as reasoned above just sketching the function and taking into account the behavior for big values of $x$. To be sure that there is no other change of sign you must prove that g' is monotone after the third point of intersection. Just after this last point, the graph of $2^x$ can easily be seen over $x^{12}$ and both subfunctions are monotone along with their derivatives: since $2^x>x^{12}\Rightarrow \log(2)2^x>12x^{11}$ which means g'(x)=12x^{11}-\log(2)2^x is indeed monotone afterwards and therefore there is no fourth intersection.