Here's some hints:

For both questions the Chinese Remainder Theorem is helpful (2019 = 3*673 and 2303 = 47*49 as you noticed).

For 2) 673 is prime so that means we can apply Fermat's Little Theorem (FLT), which helps.

For 6) we can use FLT for (mod 47). For (mod 49) remember we can use the Euler's Totient Function and the extension to FLT. Since phi(49) = 49*(6/7) = 42, remember we know that if gcd(a, 49) = 1 then $a^{\phi(49)} = a^42 \equiv 1 \pmod{49}$.

Note it is not as efficient, but it is also possible to do more direct calculations since the power isn't too large. For example, since $47\equiv -2 \pmod{49}$, we know that$$47^8 \equiv (-2)^8 \equiv 256 \equiv  11 \pmod{49}.$$ Then, for example,$$47^{43} = (47^8)^5 \cdot 47^3 \equiv 11^5\cdot (-2)^3 \pmod{49}$$and you can continue from there.

Hope these hints help!