Step 1: The nonzero residues modulo a prime \(p\) form a cyclic group \((\mathbb{Z}/p\mathbb{Z})^\times\) of order \(p-1\) under multiplication. Statement (I) claims \(a^k \equiv b^k \pmod p\) forces \(a \equiv b \pmod p\); writing \(c = ab^{-1}\), it claims \(c^k \equiv 1 \pmod p\) forces \(c \equiv 1 \pmod p\). This holds only when \(\gcd(k, p-1) = 1\); in general the subgroup of \(k\)-th roots of unity has order \(\gcd(k, p-1)\), which can exceed \(1\).
Step 2: With \(p = 7\) the group has order \(6\). Choosing \(k = 3\), \(\gcd(3,6) = 3\), so there are 3 distinct cube roots of unity mod 7, namely \(\{1, 2, 4\}\), since \(2^3 = 8 \equiv 1\) and \(4^3 = 64 \equiv 1\). Picking the nontrivial root \(c = 2\) and \(b = 1\) gives \(a = cb = 2\); then \(a^3 \equiv 1 \equiv b^3 \pmod 7\) but \(a \neq b\). Statement (I) is false.
Step 3: For statement (II), Fermat's Little Theorem gives \(a^{p-1} \equiv 1 \pmod p\) for any \(a\) coprime to \(p\), so the inverse of \(a\) is \(x \equiv a^{p-2} \pmod p\). Statement (II) instead asserts \(a \equiv a^{p-2} \pmod p\), which reduces to \(a^{p-3} \equiv 1 \pmod p\), meaning the multiplicative order of \(a\) modulo \(p\) must divide \(p-3\). Since the order of \(a\) can be any divisor of \(p-1\), and need not divide \(p-3\), the claim fails in general.
Step 4: Using \(a = 2, p = 7\): the order of \(2\) is \(3\), which does not divide \(p - 3 = 4\), so \(2^{p-3} = 2^4 = 16 \equiv 2 \not\equiv 1 \pmod 7\), confirming \(a \not\equiv a^{p-2} \pmod 7\) here as well. Statement (II) is false.
Step 5: Both statements are false.
\[\boxed{\text{Both (I) and (II)}}\]