Recall that if
$$n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$$
where $p_1 < p_2 < \cdots < p_k$ are prime numbers and $\alpha_1, \alpha_2, \ldots, \alpha_k$ are positive integers, then the sum of the positive divisors of $n$ is
$$\sigma(n) = \prod_{i=1}^{k} \left( \sum_{j=0}^{\alpha_i} p_i^j \right )$$

a. Let $n$ be a square-free perfect number. In this case, $n = p_1 p_2 \cdots p_k$, for some prime numbers $p_1 < p_2 < \ldots < p_k$, and
$$2 p_1 p_2 \cdots p_k = (p_1 + 1)(p_2 + 1) \cdots (p_k + 1)$$
If $p_1 \neq 2$ then $p_1 + 1, p_2 + 1, \ldots, p_k + 1$ are all even numbers while $p_1 p_2 \cdots p_k$ is an odd number. We deduce that $k = 1$ and $2 p_1 = p_1 + 1$
which is impossible.
If $p_1 = 2$ then $p_1 + 1 = 3 = p_2$ and if $k > 2$, we have
$$p_3 p_4 \cdots p_k = (p_3 + 1)(p_4 + 1) \cdots (p_k + 1),$$
which is impossible since both sides have different parity. Hence $k = 2$ and $n = 6$.
This proves that the only square-free perfect number is $6$.

b. Let $n$ be a perfect square. There exist prime numbers $p_1 < p_2 < \ldots < p_k$ and positive integers $\alpha_1, \alpha_2, \ldots, \alpha_k$, such that
$$n = p_1^{2 \alpha_1} p_2^{2 \alpha_2} \cdots p_k^{2 \alpha_k}$$
Because $\sum_{j=0}^{2 \alpha_i} p_i^j$ is an odd number, for $i = 1, 2, \ldots, k$, the sum $\sigma(n)$ of the positive divisors of $n$ is odd while $2n$ is even. Hence $n$ cannot be a perfect number.