Just Root Itself

For the four parts part we see there might be a solution only iff , $\displaystyle \begin{cases} x^{x^{x^{\cdots}}}=1 \to x=1 \\ x^{x^{x^{\cdots}}}=2 \to x=2^{\frac{1}{2}} \\ x^{x^{x^{\cdots}}}=3 \to x=3^{\frac{1}{3}} \\ x^{x^{x^{\cdots}}}=4 \to x=4^{\frac{1}{4}}\end{cases}$

Now due to convergence issues we must determine which of these infinite power towers converge to expected values,

$\text{Case 1 :}$

The sequence $x_0=1$ & $x_{n+1}=1^{x_n}$ represents the power tower. The exponential growth function $\displaystyle f(x)=1$ has only one fixed point $x=1$ and we see it converges to 1 as,

$\displaystyle f(x)=x \implies x=1$ is the only solution which is exactly the fixed point.

$\text{Case 2 :}$

The sequence $\displaystyle x_0=\sqrt{2}$ & $\displaystyle x_{n+1}=2^{\frac{x_{n}}{2}}$ represents the second recurrence. Similarly defining an exponential growth function $\displaystyle f(x)=2^{\frac{x}{2}}$ We observe there is a fixed point $\displaystyle x=2$ and $x=4$ where $f(x)=x$ and,

$\displaystyle f(x)=x$

$\displaystyle 2^{\frac{x}{2}}=x$

$\displaystyle x = -\frac{2W(-\frac{ln2}{2})}{ln2}$ where $\mathbf{W(.)}$ denotes Lambert W function .

Taking the principal branch $\displaystyle W_0(x)$ we see $\displaystyle a=1.99 \approx 2$ is another fixed point that converges to 2 and not to $4$ since the sequence is bounded by $2$ .

Since the sequence is bounded by $a$ which again converges to the desired fixed point we see there is a solution $x\approx 1.414$

So it has a solution.

$\text{Case 3 :}$

I am not reposting the solution as this case is excellently discussed by Comrade @Otto Bretscher in this Problem The conclusion is that no solutions are there.

$\text{Case 4:}$

Proceeding like first two cases we obtain the exponential growth function $\displaystyle f(x)=4^{\frac{x}{4}}$ where $x=4$ . Here we readily see the same power tower cannot attain the values $2,4$ at the same time. So it does not attain the value 4.

There are $\boxed{2}$ solutions.