The point of intersection of the two ellipses satisfies
$$x^2+4y^2=1$$ and $$4x^2+y^2=1$$
Subtracting,
$$3x^2-3y^2=0$$
$$x^2=y^2$$
Since $$P$$ lies in the first quadrant,
$$x=y$$
Substituting into
$$x^2+4y^2=1$$ gives $$5x^2=1$$
$$x=y=\frac1{\sqrt5}$$
Hence,
$$P=\left(\frac1{\sqrt5},\frac1{\sqrt5}\right)$$
For the ellipse
$$x^2+4y^2=1$$
differentiating implicitly,
$$2x+8y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=-\frac{x}{4y}$$
At $$P$$,
$$m_1=-\frac14$$
For the ellipse
$$4x^2+y^2=1$$
differentiating implicitly,
$$8x+2y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=-\frac{4x}{y}$$
At $$P$$,
$$m_2=-4$$
The angle between the tangents is given by
$$\tan\theta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|$$
Substituting,
$$\tan\theta=\left|\frac{-4+\frac14}{1+1}\right|$$
$$=\frac{\frac{15}{4}}{2}$$
$$=\frac{15}{8}$$
Therefore,
$$4\tan\theta=4\cdot\frac{15}{8}$$
$$=\frac{15}{2}$$
Hence,
$$\boxed{\frac{15}{2}}$$
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