Write $(12)(13)(14)$ in the form $\alpha(12)$. Where $\alpha$ is in $A_4$. The solution to this problem says solving $(12)(13)(14) = \alpha(12)$ for $\alpha$ we have $\alpha = (12)(13)(14)(12)$.
How come I can't solve it like this: $(12)(13)(14)(21) = \alpha(12)(21) = \alpha$
Write $(1234)(12)(23)$, in the form $\alpha(1234)$. Where $\alpha$ is in $A_4$. The answer is:Solving $(1234)(12)(23) = \alpha(1234)$ for $\alpha$ we have $\alpha = (234)$.
Why can't I solve it like this $(1234)(12)(23)(4321) = \alpha(1234)(4321)$, so $(1234)(12)(23)(4321) = \alpha$
You can do that, your solution works perfectly, though you might want to evaluate them further.
Note also that a transposition is self inverse, i.e. $(21)=(12)^{-1}=(12)$.
Calculating $(1234)(12)(23)(4321)$, i.e. applying from right to left: $$1\mapsto4\mapsto1 \\ 2\mapsto1\mapsto2\mapsto3\\ 3\mapsto2\mapsto3\mapsto4\\ 4\mapsto3\mapsto2\mapsto1\mapsto2$$ it will indeed give the cycle $(234)=(342)=(423)$ as the result.