1.\(\begin{array}{l}2{x^2} + 3xy + {y^2} + 5x + 3y = 11\\ \Leftrightarrow (x + y + 2)(2x + y + 1) = 13\end{array}\) 

Vì x, y nguyên nên (x;y) = (13; –14); (–12; 23); (–11; 8); (13; –28)

2.\(4{a^2} - 2ab + {b^2} = 4a + 2b \Leftrightarrow {a^2} - a.\frac{b}{2} + {\left( {\frac{b}{2}} \right)^2} = a + \frac{b}{2}\)

Đặt \(m = \frac{b}{2}\), ta có \({a^2} - am + {m^2} = a + m{\rm{  }}\left( 1 \right)\) và \(P = 253\left( {2a + b} \right) = 506\left( {a + \frac{b}{2}} \right) = 506\left( {a + m} \right)\)

\[\left( 1 \right) \Leftrightarrow {\left( {a + m} \right)^2} - 3{\rm{am}} = a + m \Leftrightarrow {\left( {a + m} \right)^2} = \left( {a + m} \right) + 3{\rm{am}}\].

Chứng minh được  \({\left( {a + m} \right)^2} \ge 4{\rm{am}} \Leftrightarrow am \le \frac{{{{\left( {a + m} \right)}^2}}}{4}\).

Suy ra \({\left( {a + m} \right)^2} = \left( {a + m} \right) + 3{\rm{am}} \le \left( {a + m} \right) + \frac{3}{4}{\left( {a + m} \right)^2} \Leftrightarrow \frac{1}{4}{\left( {a + m} \right)^2} \le a + m\)

\[ \Leftrightarrow \left( {a + m} \right)\left[ {\left( {a + m} \right) - 4} \right] \le 0 \Leftrightarrow 0 \le a + m \le 4\].

Do đó \(0 \le P \le 4.506 \Leftrightarrow 0 \le P \le 2024\).

Vậy \(\,\max P = 2024 \Leftrightarrow a = m = 2 \Leftrightarrow \left\{ \begin{array}{l}a = 2\\b = 4\end{array} \right.\).