Vectors $3\overrightarrow{\mathrm{a}}-5\overrightarrow{\mathrm{b}}$ and $2\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$ are mutually perpendicular. If $\overrightarrow{\mathrm{a}}+4\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$ are also mutually perpendicular, then cosine of the angle between $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$ is equal to

$$\begin{gathered} \left(3\overline{\mathrm{a}}-5\overline{\mathrm{b}}\right)·(2\overline{\mathrm{a}}+\overline{\mathrm{b}})=0 \\ 3{\mathrm{a}}^2-7\overline{\mathrm{a}}·\overline{\mathrm{b}}-5{\mathrm{b}}^2=0 -\left(1\right) \\ (\overline{\mathrm{a}}+4\overline{\mathrm{b}})·(\overline{\mathrm{b}}-\overline{\mathrm{a}})=0 \\ -{\mathrm{a}}^2-3\overline{\mathrm{a}}·\overline{\mathrm{b}}+4{\mathrm{b}}^2=0 -\left(2\right) \end{gathered}$$

$$\begin{gathered} 3{\mathrm{a}}^2-7\overline{\mathrm{a}}·\overline{\mathrm{b}}-5{\mathrm{b}}^2=0 \\ \underline{-3{\mathrm{a}}^2-9\overline{\mathrm{a}}·\overline{\mathrm{b}}+12{\mathrm{b}}^2=0} \\ -16 \overline{\mathrm{a}}·\overline{\mathrm{b}}+7{\mathrm{b}}^2=0 \\ \overline{\mathrm{a}}·\overline{\mathrm{b}}=\frac{7}{16}{\mathrm{b}}^2 \\ 3{\mathrm{a}}^2-\frac{49}{16}{\mathrm{b}}^2-5{\mathrm{b}}^2=0 \\ 3{\mathrm{a}}^2=\frac{129}{16}{\mathrm{b}}^2 \Rightarrow {\mathrm{a}}^2=\frac{43}{16}{\mathrm{b}}^2 \\ \mathrm{as} \cos \mathrm{\theta }=\frac{\overline{\mathrm{a}}·\overline{\mathrm{b}}}{\left|\overline{\mathrm{a}}\right| \left|\overline{\mathrm{b}}\right|}=\frac{{\displaystyle \frac{7}{16}}{\mathrm{b}}^2}{{\displaystyle \frac{\sqrt{43}}{4}}\mathrm{b}·\mathrm{b}}=\frac{7}{4\sqrt{43}} \end{gathered}$$