We have $$2\sin B=\sin A\tag1$$$$\implies \cos^2B=\frac{3+\cos^2A}{4}\tag2$$$$\implies b=\frac a2\tag3$$
You already have
$$\cos(A-B)=\frac 45,$$
i.e.
$$\cos A\cos B+\sin A\sin B=\frac 45\tag4$$
From $(1)(4)$, we have
$$\cos A\cos B=\frac 45-\frac{\sin^2A}{2}$$
Squaring the both sides and using $(2)$ give
$$\cos^2A\cdot\frac{3+\cos^2A}{4}=\left(\frac 45-\frac{1-\cos^2A}{2}\right)^2\implies \cos^2A=\frac 15$$
from which
$$\sin A=\frac{2}{\sqrt 5},\qquad \sin B=\frac{1}{\sqrt 5}$$
follow.
Since $A\gt B$, we see that $B$ is acute. Also, from $(4)$, we have
$$\cos B=+\frac{2}{\sqrt 5},\qquad \cos A=+\frac{1}{\sqrt 5}$$
from which
$$\sin\frac A2=+\sqrt{\frac{1-\cos A}{2}}=\sqrt{\frac{\sqrt 5-1}{2\sqrt 5}}$$
and
$$\sin C=\sin(\pi-A-B)=\sin(A+B)=\sin A\cos B+\cos A\sin B=1\implies C=\frac{\pi}{2}$$
follow.
By the law of cosines,
$$\frac{a}{\sin A}=\frac{c}{\sin C}\implies c=\frac{\sqrt 5}{2}a\tag5$$
Let $D$ be the intersection point of the angle bisector of $\angle A$ with the side $BC$.
Now let us consider the area of $\triangle{ABC}$.
Since $[\triangle{ABC}]=[\triangle{ABD}]+[\triangle{ACD}]$ with $|\overline{AD}|=\frac{\sqrt 2}{3}$, we have
$$\frac{1}{2}ab=\frac{1}{2}b|\overline{AD}|\sin\frac A2+\frac 12c|\overline{AD}|\sin\frac A2$$
$$\implies \frac{1}{2}\cdot a\cdot\frac{a}{2}=\frac{1}{2}\cdot\frac a2\cdot \frac{\sqrt 2}{3}\cdot \sqrt{\frac{\sqrt 5-1}{2\sqrt 5}}+\frac 12\cdot \frac{\sqrt 5}{2}a\cdot \frac{\sqrt 2}{3}\cdot\sqrt{\frac{\sqrt 5-1}{2\sqrt 5}}$$
from which
$$\color{red}{a=\frac{2}{3}\sqrt{1+\frac{1}{\sqrt 5}}}$$
follows.
By the law of sines, we have $R=\frac{a}{2\sin A}$, so
$$\color{red}{R=\frac{\sqrt{5+\sqrt 5}}{6}}$$
