1. Given a positive integer $n$ such that the last two digits of $3^{n}$ form a two-digit prime number. Then the sum of all $n$ that satisfy this condition and do not exceed 2010 is $\qquad$ .
$-, 1.909128$.
Considering $3^{k}(k=1,2, \cdots)$ modulo 100, the remainders are sequentially $3,9,27,81,43,29,87,61,83,49,47,41$, $23,69,7,21,63,89,67,1 ; 3,9, \cdots$,
where $43,29,61,83,47,41,23,89,67$ are two-digit primes, corresponding to
$$
k \equiv 5,6,8,9,11,12,13,18,19(\bmod 20) .
$$

Thus, the sum of all $n$ that satisfy the condition is
$$
\begin{array}{l}
\sum_{i=0}^{9}(9 \times 20 i+5+6+8+9+11+12+13+18+19)+ \\
\quad 2005+2006+2008+2009 \\
=909128 .
\end{array}
$$