**Solution to 1:**

Answer: 5

Solution: We can use modular arithmetic and express our remainder as [tex]r = 3^{1337} \pmod{7}[/tex]. Since 3 is coprime to 7, we can use Fermat's Little Theorem, and we find that [tex]3^6 \equiv 1 \pmod{7}[/tex]. We see that [tex]6 | 1332[/tex] since 1332 is even and 1+3+3+2=9. It follows that that for some integer k, we have [tex]r = 3^{1337} = 3^{6k + 5} = 3^{6k} * 3^{5} \equiv 1 * 3^{5} \pmod{7}[/tex], greatly simplifying the problem.

For the last step,

[tex]r = 3^{5} = \frac{3^{6}}{3} \equiv \frac{1}{3} \pmod{7}[/tex]. Thus [tex]3r \equiv 1 \pmod{7}[/tex]. By observation we see that r=5 satisfies this linear congruence, so the remainder is 5.

This looks long because I wrote out all the steps, but it can be done in the head pretty fast. Another way to do this is to make a table of remainders for small powers until it repeats. Since this table will be periodic, you can predict the remainder for larger powers. This alternate way doesn't require much knowledge of number theory.