Problem: $O$ is the circumcenter of $\triangle ABC$, which is not a right triangle. $$\frac{| AB \cdot CO |}{|AC \cdot BO|} = \frac{|AB \cdot BO|}{|AC \cdot CO|} = 3$$. Find $\tan A$. Here $AB \cdot CO$ represents the dot product of vector $\overrightarrow{AB}$ and $\overrightarrow{CO}$
My thoughts: WLOG let the radius of the excircle be $1$. Let $AB = c$, $BC = a$, $AC = b$. Then $AB \cdot CO = c \cdot 1 \cdot \cos( 90^{\circ} + \angle B- \angle A ) = -c \cdot \sin( \angle B- \angle A)$
$AC \cdot BO = -b \cdot \sin( \angle C- \angle A)$
$AB \cdot BO = c \sin C$ $AC \cdot CO = b \sin B$
But none of these look very trivial to solve..
Let us record as in the OP the values for the scalar products, then start the computation using the formulas $a=2R\sin A$ (and the similar ones for $b,c$) to express the appearing sines, and $b^2+c^2-a^2=2bc\cos A$ (and the similar ones) to express the appearing cosines. $$ \begin{aligned} |AB\cdot CO| &= AB\cdot CO\cdot\cos(\widehat{AB,CO})=cR\; \cos\left(\frac \pi2-A+B\right) =cR\sin(A-B)\ , \\ |AC\cdot BO| &= AC\cdot BO\cdot\cos(\widehat{AC,BO})=bR\; \cos\left(\frac \pi2-A+C\right) =bR\sin(A-C)\ , \\[4mm] |AB\cdot BO| &= AB\cdot BO\cdot\cos(\widehat{AB,BO})=cR\; \cos\left(\frac \pi2-C\right) =cR\sin C=2R^2\sin^2 C\ , \\ |AC\cdot CO| &= AC\cdot CO\cdot\cos(\widehat{AC,CO})=bR\; \cos\left(\frac \pi2-B\right) =bR\sin B=2R^2\sin^2 B\ . \end{aligned} $$ From the last two equalities, and the given second proportion, we get $$ 3=\frac{|AB\cdot BO|}{|AC\cdot CO|} =\frac{\sin^2 C}{\sin^2 B}=\frac {c^2}{b^2}\ . $$ So $c=b\sqrt 3$. We may want to norm $b=2$, so then $c=2\sqrt 3$. Let us use the first condition... $$ \begin{aligned} \pm 3 &= \frac{|AB\cdot CO|}{|AC\cdot BO|} \\ &=\frac{cR\sin(A-B)}{bR\sin(A-C)} =\frac cb\cdot\frac{\sin A\cos B-\cos A\sin B}{\sin A\cos C-\cos A\sin C} \\[2mm] &= \frac cb\cdot \frac {\displaystyle a\cdot\frac {a^2+c^2-b^2}{2ac} - \frac {b^2+c^2-a^2}{2bc} \cdot b} {\displaystyle a\cdot\frac {a^2+b^2-c^2}{2ab} - \frac {b^2+c^2-a^2}{2bc} \cdot c} \\[2mm] &= \frac {(a^2+c^2-b^2) - (b^2+c^2-a^2)} {(a^2+b^2-c^2) - (b^2+c^2-a^2)} \\[2mm] &= \frac {a^2-b^2} {a^2-c^2}\ . \end{aligned} $$ Recall the norming making $b^2=4$, $c^2=3b^2=12$. The two chances for $a^2$ are now $10$ and $16=4+12$, where the above becomes $-3=\frac{10-4}{10-12}$ and $+3=\frac{16-4}{16-12}$. The second case gives rise to a triangle with $A=90^\circ$, which was excluded in the title. So we have to work with the triangle with sides: $$ \color{blue}{ \boxed{\qquad a=\sqrt{10}\ ,\ b=2\ ,\ c=2\sqrt 3=\sqrt{12}\ .\qquad} } $$ Let us compute $\tan A$ by explicitly computing $\cos A$, $S$, $R$, $\sin A$ in this order: $$ \begin{aligned} \cos A&=\frac{b^2+c^2-a^2}{2bc}=\frac{4+12-10}{2\cdot 2\cdot 2\sqrt 3}=\frac 6{8\sqrt 3}=\color{blue}{\frac{\sqrt 3}4}\ , \\[2mm] S^2&=s(s-a)(s-b)(s-c)\qquad\text{(Heron)} \\ &=\frac 1{16}(a+b+c)(b+c-a)(a+c-b)(a+b-c) =\frac 1{16}\Big((b+c)^2-a^2\Big)\Big(a^2-(b-c)^2\Big) \\ &=\frac 1{16}\Big(2bc + (b^2+c^2-a^2)\Big)\Big(2bc - (b^2+c^2-a^2)\Big) =\frac 1{16}\Big(4b^2c^2 -(b^2+c^2-a^2)^2\Big) \\ &=\frac 1{16}\Big(4\cdot 4\cdot 12 -6^2\Big) =\frac{39}4\ , \\ S&=\frac {\sqrt{39}}2\ , \\[2mm] R&=\frac{abc}{4S}=2\sqrt \frac{10}{13}\ , \\[2mm] \sin A&=\frac a{2R}=\frac {\sqrt {10}}{4\sqrt {10/13}} =\color{blue}{\frac{\sqrt{13}}4}\ , \\ \tan A&=\frac{\sin A}{\cos A}=\frac{\sqrt{13}/4}{\sqrt 3/4} =\color{blue}{\boxed{\ \sqrt{\frac{13}3}}\ }\ . \end{aligned} $$ $\square$