Given that [tex]X\sim\mathrm{Unif}(0,5)[/tex], it has PDF
[tex]f_X(x)=\begin{cases}\dfrac15&\text{for }0\le x\l5\\\\0&\text{otherwise}\end{cases}[/tex]
and thus CDF
[tex]F_X(x)=\mathbb P(X\le x)=\begin{cases}0&\text{for }x<0\\\\\dfrac x5&\text{for }0\le x<5\\\\1&\text{for }x\ge5\end{cases}[/tex]
The 25th percentile is the value [tex]X=k[/tex] such that [tex]\mathbb P(X\le k)=0.25[/tex]. We have
[tex]\mathbb P(X\le k)=\dfrac k5=0.25\implies k=1.25[/tex]