1. Let the angle between the unit vectors $\vec a$ and $\vec b$ be $\theta$.

   Since $|\vec a|=|\vec b|=1$, we use the standard formulas: $|\vec a+\vec b|=\sqrt{|\vec a|^2+|\vec b|^2+2\vec a\cdot\vec b}$ $|\vec a-\vec b|=\sqrt{|\vec a|^2+|\vec b|^2-2\vec a\cdot\vec b}$

   Also, $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=\cos\theta.$

2. Therefore, $|\vec a+\vec b|=\sqrt{2+2\cos\theta}$ $|\vec a-\vec b|=\sqrt{2-2\cos\theta}.$

3. Simplify these using half-angle identities: $2+2\cos\theta=4\cos^2\frac\theta2,$ $2-2\cos\theta=4\sin^2\frac\theta2.$

   Hence,

   \qquad |\vec a-\vec b|=2\sin\frac\theta2,$$ where $0\le \theta\le \pi$, so both are nonnegative.

4. The given expression becomes

   =\sqrt3\cdot 2\cos\frac\theta2+2\sin\frac\theta2.$$ Let $$x=\frac\theta2.$$ Then $x\in[0,\pi/2]$, and $$E=2\bigl(\sqrt3\cos x+\sin x\bigr).$$

5. Now maximize $\sqrt3\cos x+\sin x.$

   Write it in the form $A\cos x+B\sin x\le \sqrt{A^2+B^2}.$

   Here $A=\sqrt3$, $B=1$, so $\sqrt{A^2+B^2}=\sqrt{3+1}=2.$

   Therefore, $\sqrt3\cos x+\sin x\le 2,$ and hence $E\le 2\cdot 2=4.$

6. This maximum is attained when $\cos x=\frac{\sqrt3}{2},\qquad \sin x=\frac12,$ i.e. $x=\frac\pi6$.

   Then $\theta=2x=\frac\pi3$, and indeed $E=4.$

Therefore, the greatest value is $\boxed{4}.$