The rigorous way to do this would be to compute a few integrals, but less work is usually better. Each of these probabilities correspond to the areas of rectangles. Each rectangle will have a height of 1, and the probability of interest will tell you everything you need to know about their lengths.
[tex]\mathbb P(0\le X\le0.4)\implies \text{length}=0.4\implies\mathbb P(0\le X\le0.4)=0.4\times1=0.4[/tex]
[tex]\mathbb P(0.4\le X\le1)\implies \text{length}=0.6\implies\mathbb P(0\le X\le0.4)=0.6[/tex]
Alternatively, you can use the fact that [tex]\mathbb P(0.4\le X\le 1)=1-\mathbb P(0\le X\le 0.4)[/tex] to get the same result.
[tex]\mathbb P(X>0.6)=\mathbb P(0.6<X\le 1)\implies \text{length}=0.4\implies\mathbb P(X>0.6)=0.4[/tex]
[tex]\mathbb P(X\le0.6)\implies \text{length}=0.6\implies\mathbb P(X\le0.6)=0.6[/tex]
[tex]\mathbb P(0.23\le X\le0.76)\implies \text{length}=0.76-0.23=0.53\implies\mathbb P(0.23\le X\le0.76)=0.53[/tex]
