a. By the Euler-Fermat theorem, since $\varphi(10) = 4$ and $4n + k \equiv k \pmod{4}$, we have
$$a^{4n+k} \equiv a^k \pmod{10}$$
for any positive integers $a$, $n$, $k$. This means $a^{4n+k}$ and $a^k$ have the same rightmost digit.

b. The rightmost digit is $0$.
By part (a), since $1999 \equiv 3 \pmod{4}$, we have
$$2^{1999} + 7^{1999} + 9^{1999} \equiv 2^3 + 7^3 + 9^3 \equiv 8 + 3 + 9 \equiv 0 \pmod{10}.$$
This shows the rightmost digit of the given number is $0$.