1. From the condition, we have $\frac{\sqrt{3}}{\sqrt{5}} b+c=-\frac{\sqrt{2}}{\sqrt{5}} a$. Let
$$y=a x^{2}+b x+c .$$

When $x=\frac{\sqrt{3}}{\sqrt{5}}$, we have
$$y_{1}=\frac{3}{5} a+\frac{\sqrt{3}}{\sqrt{5}} b+c=\frac{3-\sqrt{10}}{5} a \text {. }$$

When $x=1$, we have
$$\begin{array}{l} y_{2}=a+b+c=a+b+c-\frac{1}{\sqrt{3}}(\sqrt{2} a+\sqrt{3} b+\sqrt{5} c) \\ =\frac{a}{\sqrt{3}}\left[(\sqrt{3}-\sqrt{2})-\frac{c}{a}(\sqrt{5}-\sqrt{3})\right] . \end{array}$$

Since $3-\sqrt{10}<0, \sqrt{5}-\sqrt{3}>0$, and $-\frac{c}{a}>0$, then
$$y_{1} y_{2}=\frac{3-\sqrt{10}}{5 \sqrt{3}}\left[(\sqrt{3}-\sqrt{2}) a^{2}-\frac{c}{a}(\sqrt{5}-\sqrt{3}) a^{2}\right]>\frac{3}{4}, \text{ hence the equation has a root greater than } \frac{3}{4} \text{ and less than 1.}$$