Why an alternative magma need not even be power-associative? I don't understand well this situation

Any associative magma (i.e., a semigroup) is alternative. More generally, a magma in which every pair of elements generates an associative submagma must be alternative. The converse, however, is not true, in contrast to the situation in alternative algebras. In fact, an alternative magma need not even be power-associative.



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*Why an alternative magma need not even be power-associative ?

*When an alternative magma could be, instead, always any associative magma ?



power associativity is a property of a binary operation which is a weak form of associativity.

So in opposite direction should we have a more strong form of associativity to start from an alternative magma and return to any associative magma (that is  for definition, alternative) ?

I guess an answer to your first question can be to provide a counter-example.
Consider the magma with the following multiplication table:

It is easy to check that it is an alternative magma.
Indeed, in order to see that it satisfies
$$(x \cdot x) \cdot y = x \cdot (x \cdot y)$$
and $$x \cdot (y \cdot y) = (x \cdot y) \cdot y,$$
we don't need to check with the element $1$, since it is neutral and yields trivial identities.
Also, this magma is commutative, and so the cases in which $x$ and $y$ are the same are also trivial.
So the only relevant cases to check are the pairs $(a,b)$, $(a,c)$ and $(b,c)$, all of these satisfying the required identities.
Now in this page you can see a proof that alternative magmas have well defined powers up to the fifth, that is, they satisfy $$x^2 \cdot x = x \cdot x^2,$$
$$x^3 \cdot x = x^2 \cdot x^2 = x \cdot x^3,$$ and 
$$x^4 \cdot x = x^3 \cdot x^2 = x^2 \cdot x^3 = x \cdot x^4,$$
and so we can use exponents up to $5$ in our computations.
Now, $a^2 = b$, whence $a^3 = ba = c$, yielding
$$a^3 \cdot a^3 = c^2 = 1,$$
while, for example.
$$a^2 \cdot a^4 = b \cdot b^2 = b \cdot a = c,$$
and so $a^6$ is not well defined, and therefore the magma is not power-associative.
I don't understand the content of your second question...
