A majority of numbers are squarefree. More precisely, you can show that there is a constant $c<\frac{1}{2}$ such that for each positive integer $n$, less than $cn$ of the numbers $1$ through $n$ are nonsquarefree.
One way to get a rough bound for $c$ is to observe that at most $n/4$ of the numbers from $1$ to $n$ are divisible by $4$, at most $n/9$ divisible by $9$, at most $n/25$ divisible by $25$, etc. If we add up all these inequalities, there are less than $(1/4+1/9+1/25+\cdots)n\lt .46n$ nonsquarefree numbers from $1$ to $n$. The sum $\displaystyle{\sum_{p\text{ prime}}\frac{1}{p^2}=.4522474200410654985065...}$ is the prime zeta function evaluated at $2$.
So you can take $c=.46$. For $n\geq 13$, $\lfloor n/2\rfloor \gt .46n$, so not all of the $\lfloor n/2\rfloor$ pairwise disjoint sets $\{1,n-1\},\{2,n-2\},\ldots,\{\lfloor n/2\rfloor,n-\lfloor n/2\rfloor\}$ can contain a nonsquarefree number.
No. For example, 32 is not the sum of an odd prime and a safe prime. This is because the only safe primes smaller than 32 are 5, 7, 11, and 23, and we have:
$\begin{align*}
32&= 5 + 27\\
32&= 7 + 25\\
32&= 11 + 21\\
32&= 23 + 9\end{align*}$
I whipped up an inefficient perl program to calculate counterexamples, which include: 32, 56, 92, 98, 122, 128, 140, 152, 176, 194, 212, 224, 242, 254, 260, 272, 296, 302, 308, 326, 332, 368, 392, 398, 410, 422, 434, 452, 458, 476, 488, 500, 512, 518, 524, 536, 542, 560, 572, 596, 602, 632, 644, 656, 662, 674, 686, 692, 704, 710, 728, 752, 770, 782, 788,
800...
Based on the heuristic justification of Goldbach's conjecture and the assumption that the primality of odd $k$ and the primality of $\dfrac{k-1}{2}$ are independent, I would conjecture that there are only a finite number of such counterexamples. The expected number of solutions should be about $\dfrac{n}{2\log(n)^3}$. The same argument applies even if both primes are safe, giving about $\dfrac{n}{2\log(n)^4}$ solutions. But the experimental evidence is not very convincing, so I wonder if there is some flaw in this argument? Also, I'm suspicious of the fact that the program outputs $32$, $128$, $512$, and eventually $2048$ as well. Is there some reason that if the sum of two odd primes is an odd power of two, neither of them can be safe?
EDIT: André Nicolas has shown in his answer to my follow-up question that there are an infinite number of exceptions to this claim.
Best Answer
All positive integers except $1$ and $4$ can be written as the sum of an odd number of primes.
$6=2+2+2$ can be written as the sum of an odd number of primes.
$7=2+2+3$ can be written as the sum of an odd number of primes.
$8=2+3+3$ can be written as the sum of an odd number of primes.
$9=3+3+3$ can be written as the sum of an odd number of primes.
By adding an even number of $2$s to these, any larger number can be written as the sum of an odd number of primes.
Looking at smaller numbers, $2$, $3$ and $5$ are equal to themselves and so are equal to the sum of an odd number of primes, since one is odd, while $1$ and $4$ cannot be written as the sum of an odd number of primes.