Take $\delta=4-2.56$, so each element in the interval $]4-\delta,4+\delta[$ is taken, by $f$, into the $0.4$-neighbourhood of $2$.
Your solution for the second part somewhat works. To show the case when $a=-1$ has no global extrema, you can do what you did: you showed that the function is neither bounded above nor below. For the $a=2$ case, you want to show that $f$ has a global extrema by considering $x=0$ and show $f(x)\geq f(0)$ for all $x$. I will argue this below and give an alternative answer for $a=-1$ as well. Although, I think graphing is the simplest way for this particular problem.
Considering only global extrema, you are correct that $a=-1$ gives the result, but any other $a$ value does not. I will provide some algebraic reasoning (no derivatives) for one to make a solution. You've narrowed our choices down to $a=-1,2$ by requiring continuity, and you did this just as I would. So we consider the global extrema condition on each of these two $a$ values. I'll call $f_1:(-\infty, 1]\to \Bbb R$ with $f_1(x):=ax^2$, and $f_2:(1,\infty)\to\Bbb R$ with $f_2(x):=a^2x-2$, so that $f(x)=f_1(x)$ for $x\leq 1$ and $f(x)=f_2(x)$ for $x>1$.
Suppose $a=2$. Then $f_1$ and $f_2$ both have positive leading coefficients. Since $f_1$ is a quadratic with positive leading coefficient it must have a global minimum (in this case it is at $x=0$, which is easy to see). Since $f_2$ is a linear function with positive leading coefficient, and it is only defined for $x>1$, it must satisfy $f_2(x)\geq f_2(1)=f(1)$. Since $f(1)>f(0)$, we conclude that $f(0)$ is the minimum value of $f$, so $x=0$ is a global minimum, and thus $f$ has a global extremum.
Suppose $a=-1$. Then $f_1$ has a negative leading coefficient, so it has a global maximum but no global minimum (indeed, $f_1(x)\to -\infty$ as $x\to -\infty$). Also, $f_2$ is linear with a positive leading coefficient, so it has a lower bound (just as in the $a=2$ case), but has no global maximum (indeed, $f_2(x)\to \infty$ as $x\to\infty$). Hence, $f$ has no global maximum or global minimum, because $f_1$ has no min and $f_2$ has no max.
Best Answer
Set $g(x)=\frac{f(x^2)-1}{x}$ with $\lim_{x\to 2}g(x)=1$ Then: $$f(x^2)=xg(x)+1$$ Passing limits: $$\lim_{x\to 2}f(x^2)=\lim_{x\to 2}(xg(x)+1)=3$$ Substituting $y=x^2$: $$\lim_{y\to 4}f(y)=3$$ So: $$\lim_{x\to 4}f(x)=3$$
I alert you that when you did $\lim_{x\to 2}f(x^2)=f(4)$ is wrong, because you don't know that the function is defined at $x=4$ and you don't know that the function is continuous.