Verified answer

1) (<==) Since gcd(3, n) = 1, we have φ(3n) = φ(3) φ(n) = 2φ(n).

(==>) Suppose n = 3^k * r, where 3 does not divide r and k > 0.

Then, φ(3n) = φ(3^(k+1) * r) = φ(3^(k+1)) φ(r) = 3^(k+1) (1 - 1/3) φ(r).

and 2φ(n) = 2φ(3^k * r) = 2 φ(3^k) φ(r) = 2 * 3^k (1 - 1/3) φ(r).

Since k > 0, these are not equal.

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2) (<==) If n = 2^k, then φ(n) = φ(2^k) = 2^k (1 - 1/2) = n/2.

(==>) Note that for any odd prime p, φ(p^k) = p^k (1 - 1/p) > (p^k)/2.

So, for an integer n with more than 1 prime factor p, φ(n) > n/2, since φ is multiplicative.

I hope this helps!