First solve the congruence [tex]13y\equiv1\pmod{330}[/tex]. Euclid's algorithm shows
330 = 25 * 13 + 5
13 = 2 * 5 + 3
5 = 1 * 3 + 2
3 = 1 * 2 + 1
=> 1 = 127 * 13 - 5 * 330
=> 127 * 13 = 1 mod 330
so that [tex]y=127[/tex] is the inverse of 13 modulo 330. Then in the original congruence, multiplying both sides by 127 twice gives
[tex]127^2\cdot13^2x\equiv127^2\cdot5^2\pmod{330}\implies x\equiv127^2\cdot5^2\equiv403,225\equiv295\pmod{330}[/tex]
Then any integer of the form [tex]x=295+330n[/tex] is a solution to the congruence, where [tex]n[/tex] is any integer.
