For the hyperbola $$H: x^2 - y^2 = 1$$ and the ellipse $$E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a > b > 0$$, let the
(1) eccentricity of E be reciprocal of the eccentricity of H, and
(2) the line $$y = \sqrt{\frac{5}{2}}x + K$$ be a common tangent of E and H.
Then $$4(a^2 + b^2)$$ is equal to

Given hyperbola

$$H:x^2-y^2=1$$

Comparing with

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$

we get

$$a^2=1,\qquad b^2=1$$

Hence eccentricity of $$H$$ is

$$e_H=\sqrt{1+\frac{b^2}{a^2}}=\sqrt2$$

Given, eccentricity of ellipse $$E$$ is reciprocal of eccentricity of $$H.$$

Therefore,

$$e_E=\frac1{\sqrt2}$$

For ellipse

$$E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$

eccentricity is

$$e_E=\sqrt{1-\frac{b^2}{a^2}}$$

Thus,

$$\sqrt{1-\frac{b^2}{a^2}}=\frac1{\sqrt2}$$

Squaring,

$$1-\frac{b^2}{a^2}=\frac12$$

$$\frac{b^2}{a^2}=\frac12$$

$$a^2=2b^2\qquad\cdots(1)$$

Now,

$$y=\sqrt{\frac52}x+K$$

is tangent to both curves.

For hyperbola

$$x^2-y^2=1,$$

the tangent with slope $$m$$ is

$$y=mx\pm\sqrt{m^2-1}$$

Here,

$$m=\sqrt{\frac52}$$

Hence,

$$K^2=m^2-1=\frac52-1=\frac32\qquad\cdots(2)$$

For ellipse

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$

the tangent with slope $$m$$ is

$$y=mx\pm\sqrt{a^2m^2+b^2}$$

Therefore,

$$K^2=a^2m^2+b^2$$

Using

$$m^2=\frac52,$$

$$K^2=\frac52a^2+b^2\qquad\cdots(3)$$

From (2) and (3),

$$\frac52a^2+b^2=\frac32$$

Using

$$a^2=2b^2,$$

$$\frac52(2b^2)+b^2=\frac32$$

$$6b^2=\frac32$$

$$b^2=\frac14$$

Hence,

$$a^2=\frac12$$

Therefore,

$$4(a^2+b^2)=4\left(\frac12+\frac14\right)$$

$$=4\cdot\frac34$$

$$=3$$

Hence, the required value is

$$\boxed{3}$$.