Calcium chloride hydrolysis
Just calculating from the head by the simplifying formula $\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{(\frac {[\ce{A-}]}{[\ce{HA}]})}$, $\mathrm{pH}$ of $\pu{1 M } \ce{CaCl2}$ is approximately 1/2( 12.6 - log 1)=6.3.
So concentration $\ce{[CaOH+]}$ is about million times smaller then $\ce{[Ca^2+]}$. Hydrolysis happens, but is fully negligible. Note that the real $\mathrm{pH}$ depends on impurities and dissolved $\ce{CO2}$.
About the calculation of respective fractions, we can use the formula for gradual dissociation of multiprotic acid:
If we formally suppose $$\ce{H2A &lt;=&gt;  HA- + H+ &lt;=&gt; A^2- + 2 H+}$$ then respective molar fractions for particular acid forms are
$$x_{\ce{H2A}} = x_{\ce{Ca^2+}} =\frac {[\ce{H+}]^2}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$x_{\ce{HA-}} = x_{\ce{Ca(OH)+}} =\frac {K_\mathrm{a1} \cdot [\ce{H+}]}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
$$x_{\ce{A^2-}} = x_{\ce{Ca(OH)2}} =\frac {K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H+}]^2 + K_\mathrm{a1} \cdot [\ce{H+}] + K_\mathrm{a1} \cdot K_\mathrm{a2}}$$
where $K_\mathrm{a1} = K_\mathrm{w} / K_{\mathrm{b1},\ce{Ca(OH)2}}$ and $K_\mathrm{a2} = K_\mathrm{w} / K_{\mathrm{b2},\ce{Ca(OH)2}}$