Step 1: Find the tangent to the first curve at \((0,0)\).
The first curve is
\[
y^3-x^2y+5y-2x=0
\]
Differentiate implicitly:
\[
3y^2\frac{dy}{dx}-(2xy+x^2\frac{dy}{dx})+5\frac{dy}{dx}-2=0
\]
Collecting terms,
\[
\left(3y^2-x^2+5\right)\frac{dy}{dx}=2xy+2
\]
Thus,
\[
\frac{dy}{dx}=
\frac{2xy+2}{3y^2-x^2+5}
\]
At
\[
(0,0),
\]
we get
\[
m_1=\frac{2}{5}
\]
Step 2: Find the tangent to the second curve at \((0,0)\).
The second curve is
\[
x^4-x^3y^2+5x+2y=0
\]
Differentiate implicitly:
\[
4x^3-\left(3x^2y^2+2x^3y\frac{dy}{dx}\right)+5+2\frac{dy}{dx}=0
\]
Collecting derivative terms,
\[
\left(2-2x^3y\right)\frac{dy}{dx}
=
-4x^3+3x^2y^2-5
\]
Thus,
\[
\frac{dy}{dx}
=
\frac{-4x^3+3x^2y^2-5}{2-2x^3y}
\]
At
\[
(0,0),
\]
we get
\[
m_2=-\frac{5}{2}
\]
Step 3: Find the angle between the tangents.
For two lines with slopes \(m_1\) and \(m_2\),
\[
\tan\theta=
\left|
\frac{m_2-m_1}{1+m_1m_2}
\right|
\]
Now,
\[
m_1m_2=
\frac{2}{5}\times \left(-\frac{5}{2}\right)
=-1
\]
Since
\[
1+m_1m_2=0,
\]
we get
\[
\tan\theta=\infty
\]
Hence,
\[
\theta=\frac{\pi}{2}
\]
Step 4: Final conclusion.
Therefore,
\[
\boxed{\frac{\pi}{2}}
\]