For $x \in (1/2,1]$, let $ F_{Z|Z \leq 1}(x) = {\rm P}(Z \leq x | Z \leq 1) = \frac{{{\rm P}(Z \le x)}}{{{\rm P}(Z \le 1)}}, $ and let $ f_{Z|Z \leq 1}(x) = \frac{{\rm d}}{{{\rm d}x}}F_{Z|Z \le 1} (x) = \frac{1}{{{\rm P}(Z \le 1)}}\frac{{\rm d}}{{{\rm d}x}}{\rm P}(Z \le x). $ Then, $ {\rm E}[Z|Z \le 1] = \int_{1/2}^1 {xf_{Z|Z \le 1} (x)\,{\rm d}x} . $ So now the problem reduces to calculating ${\rm P}(Z \leq x)$, for $x \in (1/2,1]$. This can be done using the law of total probability, conditioning on $X$, leading to $ {\rm E}[Z|Z \le 1] = 7/9. $
EDIT:
Fix $x \in (1/2,1]$. Then, by the law of total probability, $ {\rm P}(Z \leq x) = \int_0^{1/2} {{\rm P}(Z \le x|X = s)\,{\rm d}s} + \int_{1/2}^1 {{\rm P}(Z \le x|X = s)\,{\rm d}s} . $ It thus follows from the definition of $Z$ (and the independence of $X$ and $Y$) that $ {\rm P}(Z \leq x) = \int_0^{1/2} {{\rm P}( Y \le x - 1/2)\,{\rm d}s} + \int_{1/2}^1 {{\rm P}(Y \le x - s)\,{\rm d}s} . $ Now, $ \int_0^{1/2} {{\rm P}( Y \le x - 1/2)\,{\rm d}s} = \frac{1}{2}{\rm P}(Y \le x - 1/2) = \frac{{x - 1/2}}{2} = \frac{x}{2} - \frac{1}{4} $ and $ \int_{1/2}^1 {{\rm P}(Y \le x - s)\,{\rm d}s} = \int_{1/2}^x {{\rm P}(Y \le x - s)\,{\rm d}s} = \int_{1/2}^x {(x - s)\,{\rm d}s} = \frac{{x^2 }}{2} - \frac{x}{2} + \frac{1}{8}. $ Hence $ {\rm P}(Z \leq x) = \bigg(\frac{x}{2} - \frac{1}{4}\bigg) + \bigg(\frac{{x^2 }}{2} - \frac{x}{2} + \frac{1}{8}\bigg) = \frac{{x^2 }}{2} - \frac{1}{8}. $ In particular, $ {\rm P}(Z \leq 1) = \frac{3}{8}. $ Thus, $ F_{Z|Z \leq 1}(x) = \frac{8}{3}\bigg(\frac{{x^2 }}{2} - \frac{1}{8}\bigg) = \frac{{4x^2 - 1}}{3}, $ and in turn $ f_{Z|Z \leq 1}(x) = \frac{{8x}}{3}. $ Finally, $ {\rm E}[Z|Z \le 1] = \int_{1/2}^1 {xf_{Z|Z \le 1} (x)\,{\rm d}x} = \frac{8}{3}\int_{1/2}^1 {x^2 \,{\rm d}x} = \frac{8}{3}\bigg(\frac{{1 - 1/8}}{3}\bigg) = \frac{7}{9}. $