If $Tf(x)= \frac{1}{\sqrt{\pi}}\int_0^x\frac{f(t)}{\sqrt{x-t}}dt$. Show that $T:L^1[0,1]\to L^1[0,1]$ and $TTf(x) = \int_0^1 f(t)dt$
$\textbf{My attempt:}$
Suppose that $\frac{1}{\sqrt{\pi}}\chi_{[0,x]}(t)\frac{f(t)}{\sqrt{x-t}}\in L^1$, then $f\in L^1$
\begin{align} \|Tf(x)\|_{L^1} & = \int_0^1\Bigg|\frac{1}{\sqrt{\pi}}\int_0^x\frac{f(t)}{\sqrt{x-t}}dt\Bigg|dx \le \frac{1}{\sqrt{\pi}}\int_0^1\int_0^1\chi_{\{t\le x\}}(t,x)\frac{|f(t)|}{\sqrt{x-t}}dtdx\\ & = \frac{1}{\sqrt{\pi}}\int_0^1\int_0^1 \chi_{\{t\le x\}}(t,x)\frac{|f(t)|}{\sqrt{x-t}}dxdt ~~~~~~~~\text{(Fubini)}\\ & = \frac{1}{\sqrt{\pi}}\int_0^1|f(t)|\int_0^1 \frac{\chi_{\{t\le x\}}(t,x)}{\sqrt{x-t}}dxdt\\ & = \frac{1}{\sqrt{\pi}}\int_0^1|f(t)|\int_t^1 \frac{1}{\sqrt{x-t}}dxdt\\ & = \frac{1}{\sqrt{\pi}}\int_0^1|f(t)|~2~\sqrt{1-t} ~dt\\ & \le \frac{2}{\sqrt{\pi}}\int_0^1|f(t)|~dt<\infty\\ \end{align}
b) \begin{align} TTf(x) &= \frac{1}{\sqrt{\pi}}\int_0^x\frac{Tf(t)}{\sqrt{x-t}}dt = \frac{1}{\sqrt{\pi}}\int_0^x\frac{1}{\sqrt{\pi}}\int_0^x\frac{f(s)}{\sqrt{y-s}}ds\frac{1}{\sqrt{x-t}}dt\\ & =\frac{1}{\pi}\int_0^x\int_0^x\frac{f(s)}{\sqrt{y-s}}\frac{1}{\sqrt{x-t}}ds dt\\ & =\frac{1}{\pi}\int_0^x\int_0^x\frac{f(s)}{\sqrt{y-s}}\frac{1}{\sqrt{x-t}}dt ds\\ & =\frac{1}{\pi}\int_0^1f(s)\int_0^1\frac{\chi_{\{s\le x\}}\chi_{\{t\le x\}}}{\sqrt{(y-s)(x-t})}dt ds\\ & =\frac{1}{\pi}\int_0^1f(s)\int_t^x\frac{1}{\sqrt{(y-s)(x-t})}dt ds\\ & =\frac{1}{\pi}\int_0^1f(s)\int_t^x\frac{1}{\sqrt{(y-s)(x-t})}dt ds\\ \end{align}
(I stuck for the rest)
Part (a) looks fine to me (I would argue you're using Tonelli's theorem instead of Fubini's, but people often combine them).
Part (b) has many errors. For one, you end up with an expression that has $x,y,t$ as free variables instead of just $x$. Here is one solution:
$$\begin{align*}TTf(x) &= \frac{1}{\sqrt{\pi}} \int_0^x \frac{Tf(t)}{\sqrt{x-t}}\,dt \\ &= \frac{1}{\pi}\int_0^x\int_0^t \frac{f(s)}{\sqrt{(t-s)(x-t)}}\,ds\,dt \\ &= \frac{1}{\pi}\int_0^x f(s)\int_s^x \frac{dt}{\sqrt{(t-s)(x-t)}}\,ds & (\text{Fubini}) \\ &= \frac{1}{\pi}\int_0^x f(s)\int_0^1\frac{du}{\sqrt{u(1-u)}}\,ds & \left(u = \frac{t-s}{x-s}\right) \\ &= \int_0^x f(s)\,ds\end{align*}$$
The step by Fubini needs a little justification, but said justification is basically just running through the entire argument with $|f|$ instead of $f$ and noting the final value is finite since $f \in L^1[0,1]$ and $0 < x < 1$.