I don't see how the first problem is related to the second

those were the only number theory questions in a 4 problem set,, probably related as thematic, coming from the same olympiad area

notice that $3^4 -2^4 \equiv 0 \mod 13$ $3^{16}-2^{16} \equiv 0 \mod 17$ Lemma $x \equiv a \mod p$ $x \equiv a \mod q$ $\gcd(p,q)=1$ $x \equiv a \mod (pq)$ $3^{16} \equiv 2^{16} \mod 13$ $3^{16} \equiv 2^{16} \mod 17$ thus $3^{16}-2^{16} \equiv 0 \mod 221$

how the computation work $3^a = 2^a \mod 13$ $(\frac{3}{2})^a = 1\mod 13$ where $2^-1$ is inverse of $2$ $2^-1*2 \equiv 1 \mod 13$ $2^-1 = 7 $ $21^a \equiv 1 \mod 13$ $8^a \equiv 1 \mod 13$ $8^2 \equiv -1 \mod 13$ so 8 is primitive mod 13 since $8^2 \equiv -1 \ne 1$ thus $a =4$ for the mod 17 $(\frac{3}{2})^b \equiv 1 \mod 17$ $(27)^b \equiv 1 \mod 17$ $10^b \equiv 1 \mod 17$ $10^2 \equiv -2 \mod 17$ $10^8 \equiv -1 \mod 17$ thus 10 is primitive then $b = 16$