[![enter image description here](https://i.stack.imgur.com/T88D5.jpg)](https://i.stack.imgur.com/T88D5.jpg)

Addendum:

I've written exercises with a simple box but I want to turn in the picture below style.Here's the code:

```
\documentclass[11pt]{book}
\usepackage[top=2cm, bottom=2.5cm, left=2.5cm, right=2.5cm]{geometry}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage[most]{tcolorbox}
\usepackage{eso-pic}
\usepackage{enumerate}
\usepackage{fancyhdr}
\usepackage{mathrsfs}
\usepackage{cancel}
\usepackage{xcolor}
\usepackage{graphicx}
\usepackage{pgf,tikz,pgfplots}
 \usepackage{varwidth}
\usepackage{listings}
\usepackage{pstricks-add}
\usepackage{tikz,tkz-tab}
\pgfplotsset{compat=1.15}
\usepackage{mathrsfs}
\usetikzlibrary{arrows}
\usepackage{mdframed}
\usepackage{hyperref}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{tikz,lmodern}
\begin{document}
\section{EXERCICES}
\begin{tcolorbox}[enhanced,breakable,colback=white,colframe=green!50!white,
colbacktitle=white!15!pink,
coltitle=pink!50!black,
borderline={0.5mm}{0mm}{green!15!white},
borderline={0.5mm}{0mm}{green!50!white,dashed},
attach boxed title to top center={yshift=-2mm},
boxed title style={boxrule=0.4pt},
title=EXERCICES]

\underline{\textbf{Exercice 1}}
\\
Simplifier les expressions suivantes:
\begin{enumerate}
    \item $A=\dfrac{5^3\times(3^2\times2)^{-1}}{3^{-1}\times(2^{-3}\times5)^2}$
    \item $B=\dfrac{(0.009)^{-3}\times(0.016)^2\times250}{(0.00075)^{-1}\times810^3\times30}$
    \item $C=\dfrac{(a^{-2}c)^{-4}(-b^2c)^5(a^3bc^{-1})^{-2}}{(-a^2b^{-3}c)^3b^4(a^{-5}c)^2}$
    \item $D=\dfrac{\left[\left(\dfrac23\right)^2\right]^6\times \left[\left(\dfrac35\right)^{-2}\right]^3\times\left[\left(\dfrac52\right)^2\right]^{-6}}{\left(\dfrac46\right)^6}$
\end{enumerate}
\underline{\textbf{Exercice 2}}
\begin{enumerate}
    \item Soient $a,b$ et $c$ trois nombres réels non nuls tels que $ab+bc+ca=0$.\\
Montrer que $\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}=-3$
\item On suppose maintenant que $abc=1$. Montrer que:
\\ $\dfrac{a}{ab+a+1}+\dfrac{b}{bc+c+1}+\dfrac{c}{ca+a+1}=1$
\item On suppose enfin que les réels $a,b$ et $c$ sont deux à deux distincts. Montrer que:
\\
$\dfrac{4a^2-1}{(a-b)(a-c)}+\dfrac{4b^2-1}{(b-c)(b-a)}+\dfrac{4c^2-1}{(c-a)(c-b)}=4$
\end{enumerate}
\underline{\textbf{Exercice 3}}
\\
\\
Soient $a,b;c$ et $d$ des nombres réels strictements positifs tels que $\dfrac{a}{b}=\dfrac{c}{d}$.
\begin{enumerate}
    \item 
    \begin{enumerate}
        \item Montrer que: $\dfrac{7a+8c}{7b+8d}=\dfrac{a}{b}$
        \item Montrer que: $\dfrac{a^2+b^2}{ab}=\dfrac{c^2+d^2}{cd}$
        \item Montrer que: $\sqrt{\dfrac{a^2+c^2}{b^2+d^2}}=\dfrac{a}{b}$
        \item Montrer que $\sqrt{(a+c)(b+d)}=\sqrt{ac}+\sqrt{bd}$
    \end{enumerate}
    \item 
    \begin{enumerate}
        \item Vérifier que si $\dfrac{a}{b}=\dfrac{c}{d}$ alors $\dfrac{a^2+b^2}{ac+bd}=\dfrac{ac+bd}{c^2+d^2}$.
        \item Démontrer la réciproque, à savoir: si $\dfrac{a^2+b^2}{ac+bd}=\dfrac{ac+bd}{c^2+d^2}$ alors $\dfrac{a}{b}=\dfrac{c}{d}$
        \\
    \end{enumerate}
\end{enumerate}

\underline{\textbf{Exercice 4}}
\\
Dans tout l'exercice $a,b$ et $c$ désignent des nombres réels

\end{enumerate}
\end{tcolorbox}
\end{document}

```

Here is a screenshot of what I have. What I want to do is:

1) Making the exercises into twocolumn format.

2) Writing exercises title and numbering like in the first picture or any other styled template for exercises specially with two column.

[![enter image description here](https://i.stack.imgur.com/lCkLn.jpg)](https://i.stack.imgur.com/lCkLn.jpg)