Prove properties of $\frac{\ln x}{x}$, and use them to show $(\sin x)^{\cos x} > (\cos x)^{\sin x} $ 
Given $$f(x)={\ln x \over x}, x>0$$
I) Find the monotony of $f.$
ΙΙ) Calculate the following integral: $$\int_1^2{\ln x \over x}{dx}$$
IIΙ) Find the domain of $f$ and, then, show that the equation $$3f(x)=1$$ has got exactly two positive roots$.$
IV) If $x_1,x_2 (x_1<x_2)$ the roots of question II. Show that exists $ξ\in(x_1,x_2)$ such as that $$3f(ξ)+3ξf'(ξ)=1.$$
V) Solve at $(0,{π\over 2})$ the inequality $$(\sin x)^{\cos x} > (\cos x)^{\sin x}.$$

Personal work:
I) $f$ is increasing at $(0,e]$ and decreasing at $[e,+\infty)$. Also, $e$ is the global maximum of $f$. $f(e)={1\over e}$
II) Let $u=\ln x$ hence $du=\frac {dx}{x}\iff xdu=dx$
The integral changes to $$\int_0^{\ln2} udu=\frac {u^2}{2}=\frac {(\ln 2)^2}{2}$$
III) The domain of $f$ is: $(-\infty,{1 \over e}]\cup[e,+\infty)=[e,+\infty)$
I've tried solving for $f(x)$ so this is what I got: $$3f(x)=1\iff f(x)={1\over 3}\iff{\ln x\over x}={1\over 3}\iff 3\ln x=x\iff\ln x={x\over 3}\iff e^{\ln x}=e^{x\over 3}\iff x=e^{x\over 3}.$$
And then I could get to nowhere. Since all the questions are linked to each other, if question III remains unsolved, thus, questions IV and V cannot be solved.
IV) I've thought of using Rolle's theorem since all the conditions are met. I chose Rolle over Bolzano because the equation has a derivative in it. Also, another idea would be that I find the anti-derivative of $$3f(ξ)+3ξf'(ξ)$$ and then let that be a function $g(x)$ and apply either Bolzano's theorem or Rolle's theorem.
V) I really have no idea about connecting $f(x)$ with either part of the inequality.
 A: Hint: The equation $$3f(x)=1$$ is equivalent to $$3\ln(x)-x=0$$ now define $$g(x)=3\ln(x)-x$$ for $x>0$ and use calculus.$$g'(x)=\frac{3}{x}-1$$
A: Your answers for I) and II) are fine.
For III) the domain of $f$ is $(0,+\infty)$ and, by your answer to question I), $f$ is strictly increasing in $(0,e]$ and strictly decreasing in $[e,+\infty)$.  Note that $f(e)=1/e>1/3$. Moreover 
$$f(1)=0<1/3\quad\text{and}\quad f(e^2)=2/e^2<1/3.$$ Now use the Intermediate value theorem and the fact that $f$ is strict monotone to show that the equation $f(x)=1/3$ has exactly one solution $x_1$ in $(1,e)$ and another one, $x_2$, in $(e,e^2)$. 
For IV) the idea of using Rolle's theorem is good. Let $x_1<x_2$ such that $3f(x_1)=3f(x_2)=1$ (they exists by question III) and consider the new function 
$$F(x):=x(3f(x)-1)=3xf(x)-x$$
(which is the anti-derivative of $3f(x)+3x f'(x)-1$). Then $F(x_1)=F(x_2)=0$ and by Rolle's theorem, there is $\xi\in(x_1,x_2)$ such that 
$$0=F'(\xi)=3f(\xi)+3\xi f'(\xi)-1.$$
Question V) is related to question I). Note that in $(0,\pi/2)$,
$(\sin x)^{\cos x}>(\cos x)^{\sin x}$
is equivalent to
$$f(\sin(x))=\frac{\ln(\sin(x))}{\sin(x)}>\frac{\ln(\cos(x))}{\cos(x)}=f(\cos(x)).$$
Since by I) $f$ is strictly increasing in $(0,1)$, it follows that it suffices to solve $\sin(x)>\cos(x)$.
A: HINT.- Just to solve V
$$\boxed{(\sin x)^{\cos x} > (\cos x)^{\sin x} \space;\space\space x\in \left(0,\frac{\pi}{2}\right)}$$
Let $$g(x)=(\sin x)^{\cos x}\\h(x)=(\cos x)^{\sin x}$$
We have $$\begin{cases}g(0)=0\text{ and } g\left(\frac{\pi}{2}\right)=1\\h(0)=1\text{ and } h\left(\frac{\pi}{2}\right)=0\end{cases}\Rightarrow g\text{ is increasing and } h \text{ is decreasing }$$
Since $\sin(x)=\cos(\frac{\pi}{2}-x)$ and $\cos(x)=\sin(\frac{\pi}{2}-x)$ we solve
$$(\sin x)^{\cos x} = (\cos x)^{\sin x}$$ which clearly has the solution $x=\dfrac{\pi}{4}$ so we finish with the solution
$$\color{red}{\dfrac{\pi}{4}\lt x\le \dfrac{\pi}{2}}$$
