Let’s first consider the case where $b=1$. In this case, we have $x^b=b^x=1^x=1$ for all $x$. Therefore, there is only one solution, namely $x=1$.

Now, let’s consider the case where $b\neq 1$. If we take the natural logarithm of both sides of $x^b=b^x$, we get:

$b \ln (x)=x \ln (b)$

If we define $f(x)=\frac{x\ln(b)}{b}$, then the equation above can be written as $f(x)=\ln(x)$. Note that $f(x)$ is an increasing function for $x>0$ if $b>1$, and a decreasing function for $0<x<1$ if $0<b<1$.

Therefore, for $b>1$, the graph of $f(x)$ intersects the graph of $\ln(x)$ at most twice, once in the interval $(0,1)$ and once in the interval $(1,\infty)$. Since $f(1)=\ln(b)$, the equation $f(x)=\ln(x)$ has a unique solution in $(1,\infty)$ if and only if $\ln(b)>0$, which is equivalent to $b>1$.

For $0<b<1$, the graph of $f(x)$ intersects the graph of $\ln(x)$ at most twice, once in the interval $(0,1)$ and once in the interval $(1,\infty)$. Since $f(1)=\ln(b)$, the equation $f(x)=\ln(x)$ has a unique solution in $(0,1)$ if and only if $\ln(b)<0$, which is equivalent to $0<b<1$.

Therefore, the equation $x^b=b^x$ has only one solution if $b=\boxed{1}$ or $0<b<1$.

To find the minimum of the function $y(x)=e^x-x^e$, we need to find its critical points. The critical points occur where the derivative of the function is zero or undefined.

Taking the derivative of $y(x)$ with respect to $x$, we get:

$$y'(x) = e^x – ex^{e-1}$$

To find the critical points, we set $y'(x) = 0$ and solve for $x$:

$$e^x – ex^{e-1} = 0$$

Dividing both sides by $e^x$ gives:

$$1 – \frac{x}{e} = 0$$

Therefore, the critical point is at $x=e$. To confirm that this is a minimum, we need to check the second derivative of $y(x)$:

$$y”(x) = e^x – e(e-1)x^{e-2}$$

Plugging in $x=e$ gives:

$$y”(e) = e^e – e(e-1)e^{e-2} = e^e – e^e + e = e$$

Since $y”(e) > 0$, the critical point at $x=e$ is a minimum.

Therefore, the function $y(x)=e^x-x^e$ has a minimum at $x=e$, and the minimum value is:

$$y(e) = e^e – e^e = 0$$

So the minimum value of the function is 0, which occurs at $x=e$.

We can approach this problem by factoring the given polynomial, if possible. One way to factor it is to use the Rational Root Theorem, which states that if a polynomial with integer coefficients has a rational root $\frac{p}{q}$ (in lowest terms), then $p$ must divide the constant term of the polynomial and $q$ must divide the leading coefficient.

In this case, the constant term is $-2$ and the leading coefficient is $1$, so any rational root must be of the form $\pm 1, \pm 2$. Checking these values, we find that none of them are roots of the polynomial.

Since the polynomial is of degree $4$, it must have exactly four complex roots (counting multiplicity). Let $a$, $b$, $c$, and $d$ be the four roots (not necessarily distinct) of the polynomial. Then we can write the polynomial as:

$$x^4 – 11x^3 + 5x – 2 = (x – a)(x – b)(x – c)(x – d)$$

Expanding the right-hand side, we get:

$$x^4 – (a + b + c + d)x^3 + (ab + ac + ad + bc + bd + cd)x^2 – (abc + abd + acd + bcd)x + abcd$$

Comparing coefficients with the left-hand side of the equation, we have:

\begin{align*} a + b + c + d &= 11 \ ab + ac + ad + bc + bd + cd &= 0 \ abc + abd + acd + bcd &= -5 \ abcd &= 2 \end{align*}

We can use these equations to eliminate variables and simplify the problem. For example, we can use the equation $abcd = 2$ to solve for one of the variables in terms of the other three:

$$d = \frac{2}{abc}$$

Substituting this into the equation $a + b + c + d = 11$ gives:

$$a + b + c + \frac{2}{abc} = 11$$

Multiplying both sides by $abc$ and rearranging, we get:

$$abc^2 + ab^2c + a^2bc + 2 = 11abc$$

This is a cubic equation in $abc$, which can be solved using various methods, such as factoring or the cubic formula. However, it turns out that the equation has no rational roots, which means that none of the roots of the polynomial $x^4 – 11x^3 + 5x – 2$ are rational.

Therefore, the only solutions to the equation are the four complex roots, which we can find using numerical methods such as the Newton-Raphson method or by using a computer algebra system.