\[x^{2} - (b - 1)x + b + 2 = 0\]
\[D = \left( - (b - 1) \right)^{2} - 4 \bullet (b + 2) =\]
\[= (b - 1)^{2} - 4b - 8 =\]
\[= b^{2} - 2b + 1 - 4b - 8 =\]
\[= b^{2} - 6b - 7 =\]
\[= (b - 7)(b + 7) > 0\]
\[x_{1} + x_{2} = b - 1;\]
\[x_{1}x_{2} = b + 2\]
\[x_{1}^{2} + x_{2}^{2} + 6x_{1}x_{2} = 13\]
\[\left( x_{1} + x_{2} \right)^{2} + 4x_{1}x_{2} = 13\]
\[(b - 1)^{2} + 4 \bullet (b + 2) - 13 = 0\]
\[b^{2} - 2b + 1 + 4b + 8 - 13 = 0\]
\[b^{2} + 2b - 4 = 0\]
\[D = 2^{2} - 4 \cdot 1 \cdot ( - 4) = 4 + 16 =\]
\[= 20\]
\[b_{1} = \frac{- 2 + \sqrt{20}}{2} = \frac{- 2 + 2\sqrt{5}}{2} =\]
\[= - 1 + \sqrt{5}\]
\[b_{2} = \frac{- 2 - \sqrt{20}}{2} = \frac{- 2 - 2\sqrt{5}}{2} =\]
\[= - 1 - \sqrt{5}\]
\[b = - 1 + \sqrt{5} \Longrightarrow\]
\[\Longrightarrow (b - 7)b + 1) < 0 \Longrightarrow \ \ \]
\[\Longrightarrow не\ подходит;\]
\[b = - 1 - \sqrt{5} \Longrightarrow\]
\[\Longrightarrow (b - 7)(b + 1) > 0.\]
\[Ответ:при\ b = - 1 - \sqrt{5}.\]