Reply to both at once.

For 6.28, we do know that $4^p \equiv 4 \pmod{p}$ for any prime, right? But then how can $4^p + 1 \equiv 0 \pmod{p}$ if $p > 5$?

For 6.30, you're on the right track, $A$ needs to be one less than a multiple of $p$, so really any $A\equiv -1 \pmod{p}$ should work. Hint here for A: (-1) raised to an odd power is always $\equiv -1 \pmod{p}$. Hint for B: Consider exponents that are powers of (p-1).