If $X_1,X_2$ are rv's having a joint PDF $f_{X_1,X_2}$ then:
$$P(X_2<X_1\leq\alpha)=\mathbb E\mathbf1_{X_2<X_1\leq\alpha}=\int\int\mathbf1_{x_2<x_1\leq\alpha}f_{X_1,X_2}(x_1,x_2)\;dx_1\;dx_2$$
If moreover $X_1,X_2$ are independent then this can be rewritten as:$$\cdots=\int\int\mathbf1_{x_2<x_1\leq\alpha}f_{X_1}(x_1)f_{X_2}(x_2)\;dx_1\;dx_2$$
Further we can change the order of integration and make use of equality $\mathbf1_{x_2<x_1\leq\alpha}=\mathbf1_{x_2<x_1}\mathbf1_{x_1\leq\alpha}$.
This can for instance lead to:$$\cdots=\int\mathbf1_{x_1\leq\alpha}f_{X_1}(x_1)\int\mathbf1_{x_2<x_1}f_{X_2}(x_2)\;dx_2\;dx_1=$$$$\int_{-\infty}^\alpha f_{X_1}(x_1)\int^{x_1}_{-\infty}f_{X_2}(x_2)\;dx_2\;dx_1=\int_{-\infty}^\alpha f_{X_1}(x_1)P(X_2\leq x_1)\;dx_1=\int_{-\infty}^\alpha f_{X_1}(x_1)F_{X_2}(x_1))\;dx_1$$
The same technique: $$P(\text{condition on }X_1,X_2,X_3)=\mathbb E\mathbf1_{\text{condition on }X_1,X_2,X_3}$$ can be applied to find an integral expression for $P(X_1+X_2\leq\alpha\wedge X_1<X_2)$.
addendum:
$\begin{aligned}\int\int\int\mathbf{1}_{x_{3}\leq\alpha-x_{1}}\mathbf{1}_{x_{1}<x_{2}}f_{X_{1}}\left(x_{1}\right)f_{X_{2}}\left(x_{2}\right)f_{X_{3}}\left(x_{3}\right)dx_{3}dx_{1}dx_{2} & =\int f_{X_{2}}\left(x_{2}\right)\int\mathbf{1}_{x_{1}<x_{2}}f_{X_{1}}\left(x_{1}\right)\int\mathbf{1}_{x_{3}\leq\alpha-x_{1}}f_{X_{3}}\left(x_{3}\right)dx_{3}dx_{1}dx_{2}\\
& =\int f_{X_{2}}\left(x_{2}\right)\int\mathbf{1}_{x_{1}<x_{2}}f_{X_{1}}\left(x_{1}\right)F_{X_{3}}\left(\alpha-x_{1}\right)dx_{1}dx_{2}\\
& =\int f_{X_{2}}\left(x_{2}\right)\int_{-\infty}^{x_{2}}f_{X_{1}}\left(x_{1}\right)F_{X_{3}}\left(\alpha-x_{1}\right)dx_{1}dx_{2}
\end{aligned}
$
There could possibly be a simpler argument that does not require doing the integration explicitly.
For the integration, I felt it convenient to use a change of variables.
You are looking for
$$P(X_{(1)}+X_{(2)}<X_{(3)})=6\iiint\mathbf1_{x+y<z\,,\,0<x<y<z<1}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$
Change variables $(x,y,z)\to(u,v,w)$ with $$u=x+y\,,\,v=y\,,\,w=z$$
Then, $$x+y<z\,,\,0<x<y<z<1\implies u<w\,,\,0<u-v<v<w<1$$
Therefore,
\begin{align}
\iiint\mathbf1_{x+y<z\,,\,0<x<y<z<1}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z&=\int_0^1\int_0^w\int_v^{\min(2v,w)}\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w
\\&=\int_0^1\int_0^{w/2}\int_v^{2v}\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w+\int_0^1\int_{w/2}^w\int_v^w\,\mathrm{d}u\,\mathrm{d}v\,\mathrm{d}w
\\&=\frac{1}{12}
\end{align}
Best Answer
Observe that for $x_i \in (0,1)$, if $x_1+x_2+x_3 <1$, then $x_1+x_2<1$ will automatically be satisfied. In other words, $B \subseteq A$. Thus $A \cap B^c=A\setminus B$. So the probability $P(A \cap B^c)=P(A)-P(B)=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$.