Short answer.
In general we define $a^b = e^{b\ln a}$, so $(x-4)^x = e^{x\ln(x-4)}$ is not a well defined real function for $x \leq 4$, and the given limit can't exist if we are considering only real numbers.
To answer the second question:
$$\lim_{x\to 2} (x+4)^x = \lim_{x\to 2} e^{x\ln(x+4)} = e^{\lim_{x\to 2}(x\ln(x+4))} = e^{2\ln 6} = 6^2 = 36$$
and there are no problems since $x+4$ is positive for $x > -4$ and $2> -4$.
(Really?) long answer.
Let me first explain how $a^b$ is defined when $a$ and $b$ are real numbers. I will assume for now that $a>0$ and will later explain the issues that arise if we don't make this assumption.
Fix some $a>0$. At first let $b = n$ be a positive integer. We are well aware that $$a^n = \underbrace{a\cdot a\cdot a\cdot \ldots \cdot a}_n.\tag{1}$$
We can easily extend this definition to all integers $n$ by defining $a^0 = 1$ and defining $$a^{-n} = \frac 1{a^n}\tag{2}.$$
Definitions $(1)$ and $(2)$ are very satisfying since we have the familiar rules for powers like $a^{m+n} = a^m\cdot a^n$ and $(a^m)^n = a^{mn}$, which work for any integers $m,n$.
Next, we would like to define $a^b$ when $b$ is a rational number. First, let $b = \frac 1n$ for a positive integer $n$. Now we can define $$a^{1/n} = \sqrt[n] a\tag{3}.$$
At this point we need to pause and think about what $n$-th root means. It should be a solution to the equation $x^n = a$ because we want $n$-th root to be opposite of taking $n$-th power, i.e. $(\sqrt[n]a)^n = a$. However, if you know about complex numbers, you know that there are exactly $n$ complex solutions to the equation $x^n = a$. So, if we want $n$-th root to be well defined as a function, we are forced to make a choice which solution we want. Luckily, when $a > 0$ there is always a unique positive real solution to the equation $x^n = a$. This should be familiar when you ask the question what is $\sqrt 1$. Although $1^2 = (-1)^2 = 1$, $1>0$, while $-1 < 0$, so $\sqrt 1 = 1$. This is more complicated if we ask what is $\sqrt[4] 1$, since we now have $1^4 = (-1)^4 = i^4 = (-i)^4 = 1$ (where $i$ is the imaginary unit), but only $1$ is positive real number. This is maybe a long story to justify that when $a > 0$, there is no issue in making sense of the definition $(3)$.
Now we are ready to extend our definition to rational numbers. Let $b = m/n$ where $m$ is an integer and $n$ is a positive integer. We can define $$a^{m/n} = (\sqrt[n] a)^m\tag 4$$ and you can see our rules $a^{b+c} = a^b\cdot a^c$ and $(a^b)^c = a^{bc}$ are readily satisfied, so we are happy with this.
The next step is to extend our definition to arbitrary real number $b$. The way how real numbers work is that rational numbers are dense in real numbers, which means we can obtain any real number $b$ as a limit of a sequence of rational numbers $(q_n)$. We would like that our definition of $a^b$ is continuous, i.e. $\lim_n a^{q_n} = a^{\lim_n q_n} = a^b$. However, since there are many different sequences $(q_n)$ of rational numbers that converge to the same real number $b$, to avoid discussion that our definition doesn't depend on the choice of sequence $(q_n)$, we can use the well known continuous function: $x\mapsto e^x$. How this is defined goes beyond the scope of high school but let us just assume that this function is given to us and that it satisfies the properties $e^{x+y} = e^x\cdot e^y$ and $e^0 = 1$ and $(e^b)^c = e^{bc}$. We can now define $$a^b = e^{b\ln a}\tag 5.$$ Note that this definition is reasonable, since $e^{b\ln a} = (e^{\ln a})^b = a^b$ for any rational number $b$, and since exponential function is continuous, we immediately get that $\lim_n a^{q_n} = \lim_n e^{q_n\ln a} = e^{\lim_n q_n\ln a} = e^{b\ln a}$, so our definition $(5)$ works beautifully for any real $b$ and $a>0$. It also generalizes all the previous definitions we gave.
So, for any positive real number $a$, we can define a real function $x\mapsto a^x$ by the formula $a^x = e^{x\ln a}.$
At this point, it might have become clear where the problem is when $a$ is not a positive real: $\ln a$ is not a real number! To extend definition of $a^x$ for negative real numbers $a$, we need to deal with complex numbers and go to the field of complex analysis which is way beyond the scope of high school. I will not go there, but let me add something more about the problem
The problem with $a < 0$ first arises in considering definition $(3)$. You can see that there is absolutely no issue in definitions $(1)$ and $(2)$ since we are only dealing with multiplication and division, so something like $(-2)^{-3} = -1/8$ is not problematic at all.
Let's extend discussion of $(3)$ (and $(4)$). When $n$ is an odd positive integer, there is no problem with $\sqrt[n]a$ since for $a<0$ and odd positive integer $n$, there is always unique negative real solution to the equation $x^n = a$. For example, $\sqrt[3]{-1} = -1$. But, when $n$ is even, we have a problem, what is, for example $\sqrt{-1}$? The answers could be $i$ or $-i$ and usually we make a convention that $\sqrt{-1} = i$ because that convention is very useful when you consider quadratic formula. But it is not without any issues, you might have seen the famous wrong proof
$$1 = \sqrt 1 = \sqrt{(-1)\cdot(-1)} = \sqrt{-1}\cdot \sqrt{-1} = i\cdot i = -1,$$ which tells us that a rule like $\sqrt ab = \sqrt a\cdot\sqrt b$ falls apart when we are not dealing with positive $a$ and $b$.
Furthermore, what is $\sqrt[4]{-4}$? You can check that $(1+i)^4 = (1-i)^4 = (-1+i)^4 = (-1-i)^4 = -4$. What do we choose now?
Whatever we would choose, we break the rule $(a^b)^c = a^{bc}$, for example
$$4 = ((-4)^4)^{1/4} \neq ((-4)^{\frac 14})^4 = -4.$$
To systematically deal with problems like these one needs to take definition $(5)$ and extend definition of logarithm to (almost) all complex numbers. But we are now leaving the realm of real numbers and real analysis.
