Given that $X_1,X_2,X_3 \sim_{i.i.d.} U(0,1) $, and those order statistics are $Y_1=X_{(1)},Y_2=X_{(2)},Y_3=X_{(3)}$, assuming $X_{(1)} \le X_{(2)} \le X_{(3)}$.
Suppose $Y_1 = T$ and $Z=X_{(3)}-X_{(1)}$, the probability $f_{Z,T}(z,t)$ is
$$ \begin{align} f_{Z,T}(z,t) &= f_{Y_{1},Y_{3}}(y_{1},y_{3}) \mid J((z,t)\to(y_{1},y_{3})) \mid \\ &= f_{Y_{1},Y_{3}}(t,z+t) \\ &= f_{1}(t)f_{3}(z+t) \\ &= 9(1-t)^2(z+t)^2. \end{align} $$
$f_1(\cdot),f_3(\cdot)$ is p.d.f of $Y_1$ and $Y_3$ each and
$$ \begin{align} f_{1}(y) &= \frac{3!}{2!} \times 1 \times (1-y)^2 = 3(1-y)^2, \\ f_{3}(y) &= \frac{3!}{2!} \times y^2 \times 1 = 3y^2. \end{align} $$
Therefore
$$ \begin{align} f_{Z}(z) &= 9\int_{0}^1 (1-t)^2(z+t)^2 \, dt \\ &= 9\int_{0}^1 s^2(z+(1-s))^2 \, ds \\ &= 9\left( \frac{1}{5} - \frac{1}{2}(z+1) + \frac{1}{3}(z+1)^2 \right) \\ &= 9\left( \frac{1}{3}z^2 + \frac{1}{6}z + \frac{1}{30} \right) \\ &= 3 \left( z^2 + \frac{1}{2}z + \frac{1}{10} \right). \end{align} $$
Then $$ \begin{align} E[Z] &= \int_{0}^1 zf_{Z}(z) \, dz \\ &= 3\left[ \frac{1}{4}z^4 + \frac{1}{6}z^3 + \frac{1}{20}z^2 \right]_{0}^1 \\ &= 3\frac{1}{2} \left( \frac{1}{2} +\frac{1}{3} + \frac{1}{10} \right) \\ &= \frac{7}{5}. \end{align} $$
But the answer is $E[Z] = \frac{1}{2}, V[Z] = \frac{1}{20}$.
Where did I make mistakes?
Something worth keeping in mind is the linearity of expected value. Thus, the expected value of $X_{(3)}-X_{(1)}$ is the expected value of $X_{(3)}$ minus the expected value of $X_{(1)}$. This holds even for dependent random variables like the order statistics are.
You can go through the integral calculus calculations to show this formally, but, intuitively, uniformly splitting the interval into four pieces should yield expected piece lengths of a quarter, so expected break points (the order statistic expected values) at $1/4$, $1/2$, and $3/4$. This leads to $3/4-1/4=1/2$, which is the given expected value
It’s also correct to calculate the expected value of $Z$ directly by integrating its PDF, and practicing such a calculation might be part of point of the exercise, but that seems like a mistake-prone approach. Easy integration of rectangles beats convolutions if all you care to derive is the expected value.