$10 x$ is a positive integer, the last two digits of $3^{x}$ form a prime number, then the two values of $x$ closest to 2008 are $\qquad$ and $\qquad$ .
10 Calculation shows that $3^{20} \equiv 1(\bmod 100)$, so
$$
3^{2008}=\left(3^{20}\right)^{100} \times 3^{8} \equiv 61(\bmod 100) .
$$
61 is a prime number, so $x=2008$ is a number that meets the requirement. And
$$
3^{2007}=\left(3^{20}\right)^{100} \times 3^{7} \equiv 87(\bmod 100),
$$

as well as
$$
3^{2009}=\left(3^{20}\right)^{100} \times 3^{9} \equiv 83(\bmod 100),
$$
87 is a composite number, 83 is a prime number, so $x=2009$ is another number that meets the requirement.