Division by $0$
The first question you need to ask is: What does "$a/b$" mean?
The answer is: "$a/b$ is the unique solution to the equation $bz = a$." (I'm using $z$ as the unknown, since you are using $x$ for other things).
Given that answer, let's discuss your points out of order:
(3) is perfectly fine: $0/x$, with $x\neq 0$, is the solution to $xz = 0$; the unique solution is $z=0$, so $0/x = z$. The reason it's unique is because $x\neq 0$, so the only way for the product to be $0$ is if $z$ is $0$.
In (1), by "impossible" we mean that the equation that defines it has no solutions: for something to be equal to $x/0$, with $x\neq 0$, we would need $0z = x$. But $0z=0$ for any $z$, so there are no solutions to the equation. Since there are no solutions to the equation, there is no such thing as "$x/0$". So $x/0$ does not represent any number.
In (2), the situation is a bit trickier; in terms of the defining equation, the problem here is that the equation $0z=0$ has any value of $z$ as a solution (that's what the "infinite solutions" means). Since the expression $a/b$ means "the unique solution to $bx= a$, then when $a=b=0$, you don't have a unique answer, so there is no "unique solution".
Generally speaking, we simply do not define "division by $0$". The issue is that, once you get to calculus, you are going to find situations where you have two variable quantities, $a$ and $b$, and you are considering $a/b$; and as $a$ and $b$ changes, you want to know what happens to $a/b$. In those situations, if $a$ is approaching $x$ and $b$ is approaching $y\neq 0$, then $a/b$ will approach $x/y$, no problem. If $a$ approaches $x\neq 0$, and $b$ approaches $0$, then $a/b$ does not approach anything (the "limits does not exist"). But if both $a$ and $b$ approach $0$, then you don't know what happens to $a/b$; it can exist, not exist, or approach pretty much any number. We say this kind of limit is "indeterminate". So there is a reason for separating out cases (1) and (2): very soon you will see an important qualitative difference between the first kind of "does not exist" and the second kind.
The key is to realize what a fraction $\frac ab$ really represents: $\frac ab$ is the number with the property that $\frac ab \cdot b = a$.
So, in the first case if $x \ne 0$, there is no number $\frac x0$ with the property that $\frac x0 \cdot 0 = x$ since anything times zero is zero. So $\frac x0$ is undefined. In the second, any number $y$ has the property that $y \cdot 0 = 0$, so $\frac 00$ could represent any number $y$ according to the above characterization of a fraction, so $\frac 00$ is said to be indeterminate. Finally, if $x \ne 0$ then $0$ has the property that $0 \cdot x = 0$, so $\frac 0x$ is the number $0$.
Definition of Division
For every real number a and every nonzero real number b, the quotient a$\div$b,
or $\dfrac{a}{b}$, is defined by:
$$a\div b=a \cdot \frac{1}{b}.$$
Dividing by zero would mean multiplying by the reciprocal of 0.
But 0 has no reciprocal (because 0 times any number is 0, not 1.)
Therefore, division by 0 has no meaning in the set of real numbers.
Multiplicative property of 0
Prove:
If $a$ is any real number, then $a\cdot 0 = 0$ and $0\cdot a = 0$.
Proof:
Statement _________________Reason
$0 = 0 + 0$ ______________1. Identity property of addition
$a\cdot0 = a(0 + 0)$ __________2. Multiplication property of equality
$a\cdot0 = a\cdot0 + a\cdot0$ ________3. Distributive property of mult. with respect to add.
But $a\cdot0 = a\cdot0 + 0$ _______4. Identity property of addition
$\therefore$ $a\cdot0 + a\cdot0 = a\cdot0 + 0$ ____5. Transitive property of equality
$a\cdot0 = 0$ _______________6. Subtraction property of equality
$0\cdot a = 0$ _______________7. Commutative property of multiplication
Therefore, 0 times any number is 0, not 1.
(Source: Algebra: Structure and Method Book 1)
The two cases presented in an older edition of Book 2 of the above source are:
Dividing a nonzero number by zero, violates the multiplicative property of zero and therefore the properties of the real numbers upon which it is proven, as shown above.
Dividing zero by zero, which does not violate the multiplicative property of zero, but multiplication by zero is an operation that results in zero for every real number.
If $\dfrac{a}{0}$ = c, then $a = 0\cdot c$. But $0\cdot c = 0$. Hence, if $a$ is not equal to $0$, no value of $c$ can make the statement $a = 0\cdot c$ true, while if $a = 0$, every value of $c$ will make the statement true.
Thus, $\dfrac{a}{0}$ either has no value or is indefinite in value.
This separation into two cases, one of which results in no value satisfying the multiplicative property of zero and the other resulting in an indefinite value satisfying it, gives the impression that $\dfrac 0{0}$ is allowed.
The following argument starts with the equation: $$a \cdot \frac{1}{a} = 1$$ and notes that $a\neq 0$ because $0$ times any number is $0$.
Thus, the product of $0$ and no real number equals $1$.
This further reinforces the idea that $0$ has no number that when multiplied by it equals $1$.