\[x^{2} - (b + 1)x + b + 2 = 0\]
\[D = (b + 1)^{2} - 4 \bullet (b + 2) =\]
\[= b^{2} + 2b + 1 - 4b - 8 =\]
\[= b^{2} - 2b - 7 =\]
\[x_{1} + x_{2} = b + 1;\ \ \ \ x_{1}x_{2} = b + 2\]
\[x_{1}^{2} + x_{2}^{2} + 5x_{1}x_{2} = 33\]
\[\left( x_{1} + x_{2} \right)^{2} + 3x_{1}x_{2} - 33 = 0\]
\[(b + 1)^{2} + 3 \bullet (b + 2) - 33 = 0\]
\[b^{2} + 2b + 1 + 3b + 6 - 33 = 0\]
\[b^{2} + 5b - 26 = 0\]
\[D = 5^{2} - 4 \cdot ( - 26) =\]
\[= 25 + 104 = 129\]
\[b_{1} = \frac{- 5 + \sqrt{129}}{2};\ \ \ \ \]
\[b_{2} = \frac{- 5 - \sqrt{129}}{2}\]
\[Ответ:при\ b = \frac{- 5 - \sqrt{129}}{2}.\]