Let $A \in \mathcal{B} (L_2(0,1))$ be defined by $Af(t)= tf(t)$. We want to show that $A \geq 0$ and find $A^{1/n}$.
What I have tried for the first part of the question is as follows. It is known that $A \geq 0 $ if and only if $\langle f, Af \rangle \geq 0$ for all $f \in L_2(0,1)$. Let $f \in L_2(0,1)$ and so \begin{align*} \langle f , Af \rangle &= \int_0^1 f(t) \overline{Af}(t)dt \\ & = \int_0^1 t f(t) \overline{f(t)} dt = \int_0^1 t |f(t)|^2 dt \end{align*} which is positive. Firstly I am not sure my reasoning is correct or not, and also I have no clue the second part of the question. In fact we know that since $A \geq 0$, thus, there exist a unique positive element $T$ such that $A = T^n$. However, how should we find an explicit form for that? Thanks in advance for your helps.
Your proof of $A\geq 0$ is correct. More generally, multiplication with a positive bounded function is a positive operator on $L^2$ (see below).
One way to get $A^{1/n}$ is simply to show that the multiplication by $t^{1/n}$ satisfies the defining properties of $A^{1/n}$.
More conceptually, you can prove that the map $\Phi\colon C([0,1])\to \mathcal{B}(L^2((0,1))$ that maps a function $\phi$ to the multiplication operator $\Phi(\phi)f:=\phi f$ is linear and satisfies $\Phi(\phi\psi)=\Phi(\phi)\Phi(\psi)$, $\Phi(\overline\phi)=\Phi(\phi)^\ast$. It is an immediate consequence that $\Phi(|\phi|^2)=\Phi(\phi)^\ast\Phi(\phi)$, hence positive functions are mapped to positive operators.
Moreover, it follows by induction that $\Phi(\phi^n)=\Phi(\phi)^n$ for all $n\in\mathbb{N}$. So you're left with the task to find a positive function $\phi\in C([0,1])$ that satisfies $\phi(t)^{n}=t$, and well, there is $\phi(t)=t^{1/n}$ again.