We know the equation of the tangent at $P(h,k)$ of $(x-a)^2+(y-b)^2=r^2$ is $(x-h)(h-a)+(y-k)(k-b)=0$, so the gradient is $-\frac{h-a}{k-b}$
and the parametric equation of the circle is $x=a+r\cos A,y=b+r\sin B$, so the gradient at $(a+r\cos A,b+r\sin B)$ is $-\cot A$
For $(x-3)^2+y^2=5,a=3,b=0,r=\sqrt 5,$ so, the gradient at $(3+\sqrt 5\cos B,\sqrt 5\sin B)$ is $-\cot B$
For the required circle $(x-4)^2+(y-3)^2=R^2,a=4,b=3,r=R$ so the gradient at $(4+R\cos C,3+R\sin C)$ is $-\cot C$
To be perpendicular, $-\cot C(-\cot B)=-1$ $\implies \tan B=-\cot C=\tan(\frac{\pi}2+C) \implies B=\frac{\pi}2+C$
So, $(3+\sqrt 5\cos B,\sqrt 5\sin B)$ becomes $(3-\sqrt 5\sin C,\sqrt5\cos C)$
So, $3-\sqrt5\sin C=4+R\cos C$,$\sqrt5\cos C=3+R\sin C$
or $R\cos C+\sqrt5 \sin C=-1$ and $R\sin C-\sqrt 5cos C=-3$
Squaring and adding $R^2+ 5=1^2+3^2\implies R=\sqrt5$ as $R>0$