(a)
\[
P(X\leqslant 1)=F(1)=\frac{1}{4}\Bigl[1+\ln(\frac{4}{1})\Bigr]=\frac{1}{4}(1+2\ln 2)\approx 0.59657359
\]
(b)
\[
\begin{split}
P(1\leqslant X\leqslant 3)&=F(3)-F(1)\\
&=\frac{3}{4}\Bigl[1+\ln(\frac{4}{3})\Bigr]-\frac{1}{4}\Bigl[1+\ln(\frac{4}{1})\Bigr]\\
&=\frac{3}{4}(1+2\ln 2-\ln 3)-\frac{1}{4}(1+2\ln 2)\\
&=\frac{3}{4}+\frac{3}{2}\ln 2-\frac{3}{4}\ln 3-\frac{1}{4}-\frac{1}{2}\ln 2\\
&=\frac{1}{2}+\ln 2-\frac{3}{4}\ln 3\\
&\approx 0.3691879900
\end{split}
\]
(c)
For $x < 0$ or $x > 4$, $f(x)=0$, since $F(x)$ is constant.
For $x=0$, it is clear that $F'_{-}(0)=0$ and
\[
\begin{split}
F'_{+}(0)&=\lim_{x\rightarrow 0^+}\frac{F(x)-F(0)}{x-0}\\
&=\lim_{x\rightarrow 0^+}\frac{\frac{x}{4}(1+\ln\frac{4}{x})-0}{x}\\
&=\lim_{x\rightarrow 0^+}\frac{1}{4}(1+\ln\frac{4}{x})\\
&=+\infty
\end{split}
\]
So, the derivative of the function $F(x)$ at the point $x=0$ doesn't exist.
For $x\in(0,4)$,
\[
\begin{split}
f(x)&=F'(x)=\biggl(\frac{x}{4}(1+\ln\frac{4}{x})\biggr)'\\
&=\biggl(\frac{x}{4}\biggr)'\cdot(1+\ln\frac{4}{x})+\frac{x}{4}\cdot\biggl(1+\ln\frac{4}{x}\biggr)'\\
&=\frac{1}{4}\cdot(1+\ln\frac{4}{x})+\frac{x}{4}\cdot\frac{x}{4}\cdot\biggl(\frac{4}{x}\biggr)'\\
&=\frac{1}{4}+\frac{1}{4}\ln\frac{4}{x}+\frac{x^2}{16}\cdot\frac{-4}{x^2}\\
&=\frac{1}{4}+\frac{1}{4}\ln\frac{4}{x}-\frac{1}{4}\\
&=\frac{1}{4}\ln\frac{4}{x}
\end{split}
\]
For $x=4$, it is clear that $F'_{+}(4)=0$.
\[
\begin{split}
F'_{-}(4)&=\lim_{x\rightarrow 4^-}\frac{F(x)-F(4)}{x-4}\\
&=\lim_{x\rightarrow 4^-}\frac{\frac{x}{4}(1+\ln\frac{4}{x})-1}{x-4}\\
&=\lim_{x\rightarrow 4^-}\frac{x(1+\ln\frac{4}{x})-4}{4(x-4)}\\
&=\lim_{x\rightarrow 4^-}\frac{x-4+x\ln(1+\frac{4-x}{x})}{4(x-4)}\\
&=\frac{1}{4}+\lim_{x\rightarrow 4^-}\frac{x\ln(1+\frac{4-x}{x})}{4(x-4)}\\
&=\frac{1}{4}+\lim_{x\rightarrow 4^-}\frac{\ln(1+\frac{4-x}{x})}{x-4}\\
&=\frac{1}{4}+\lim_{x\rightarrow 4^-}\frac{\frac{4-x}{x}}{x-4}\\
&=\frac{1}{4}-\lim_{x\rightarrow 4^-}\frac{1}{x}\\
&=\frac{1}{4}-\frac{1}{4}\\
&=0
\end{split}
\]
Hence, $f(4)=F'(4)=0$.
Therefore, we have
\[
f(x)=\begin{cases}
0, & x < 0,\\
\frac{1}{4}\ln\frac{4}{x}, & 0 < x < 4,\\
0, & x\geqslant 4.
\end{cases}
\]
If write $f(x)$ in decimal format, we have $f(x)=0.34657359-0.25\ln x$ for $x\in(0,4)$.
