We have that $X\sim U[0,2]$ and $Y=\min (X,1)$.
I want to find the distribution of $Y$.
We have that \begin{align*}F_Y(y)&=P(Y\leq y)=P(\min (X,1)\leq y)=P([X\leq y] \text{ or } \ [1\leq y])=P(\{X\leq y\}\cup \{1\leq y\})\\ & =P(X\leq y)+P(1\leq y)\end{align*}
Is everything correct so far?
Since $X\sim U[0,2]$ do we have that $P(X\leq y)=y$ ? What does it hold for $P(1\leq y)$ ?
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Coul you give me also a hint how we could calculate the expected value of $Y$ ?
$P\big(\{X\le y\}\cup\{1\le y\}\big)$ need not be equal to $P(X\le y)+P(1\le y)$, unless the two events are disjoint. We have $$P(X\le y)=\begin{cases}0,&y<0\\\frac{y}{2},&0\le y< 2\\1,&2\le y\end{cases}$$ Also trivially, we have $$P(1\le y)=\begin{cases}0,&y<1\\1,&y\ge 1\end{cases}$$ Then we have $\{X\le y\}\cap\{1\le y\}=\begin{cases}\emptyset,&y<1\\\{X\le y\},&y\ge 1\end{cases}$
So $$F_Y(y)=P\big(\{X\le y\}\cup\{1\le y\}\big)=P(X\le y)+P(1\le y)-P\big(\{X\le y\}\cap\{1\le y\}\big)=\begin{cases}P(X\le y)+P(1\le y)=\frac{y}{2},&y<1\\P(1\le y)=1,&y\ge 1\end{cases}$$
Thus $F_Y$ has a jump discontinuity at $y=1$ and we get $$P(Y=1)=P(Y\le 1)-P(Y<1)=F_Y(1)-\lim_{y\to 1,\,y<1}F_Y(y)=1-\frac{1}{2}=\frac{1}{2}.$$
We also have $f_Y(y)=\frac{1}{2},\,y<1$. So $$E[Y]=\int_0^1yf_Y(y)dy+1\cdot P(Y=1)=\int_0^1\frac{y}{2}dy+\frac{1}{2}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}.$$