= Solution
The independent <uniform distributions> give a constant <joint probability density> $1$ on the unit square. For $0<x<1$, integrate over $x<u<v^2$:
$$
P(V^2>U>x)=\int_x^1\int_{\sqrt u}^1dv\,du=\int_x^1(1-\sqrt u)du
=\boxed{\frac13-x+\frac23x^{3/2}}.
$$
For the <random quadratic with uniform coefficients>, real roots require its <discriminant> to be nonnegative, equivalently $U\leq V^2$. The boundary has <probability> zero, and the area under this parabola is
$$
\boxed{P(\text{real roots})=\int_0^1v^2dv=\frac13}.
$$
On that event the roots are $-V\pm\sqrt{V^2-U}$, both nonpositive. Bounding both absolute values by one therefore amounts to bounding the more negative root:
$$
V+\sqrt{V^2-U}\leq1.
$$
Since $0\leq V\leq1$, squaring the inequality is legitimate and gives $U\geq2V-1$. The admissible unit-square region is $\max(0,2v-1)\leq u\leq v^2$. Its <probability> is
$$
\int_0^{1/2}v^2dv+\int_{1/2}^1(v^2-2v+1)dv=\frac1{24}+\frac1{24}=\frac1{12}.
$$
Dividing by the real-root <probability> gives the requested <conditional probability>:
$$
\boxed{P(|R_1|\leq1,|R_2|\leq1\mid\text{real roots})=\frac14}.
$$
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