Lời giải:

• Đặt \( t = \sin^2 x \) \(\Rightarrow 0 \leq t \leq 1\)
• \( y = \frac{\cos^2 x + m}{2 \sin^2 x - m} = \frac{-\sin^2 x + 1 + m}{2 \sin^2 x - m} = \frac{-t + 1 + m}{2t - m} = f(t) \)
• \( y' = \frac{-2 - m}{(2t - m)^2} \)
• \(\max y = \max\limits_{[0, 1]} f(t) = 2\)

\(\Leftrightarrow \left[ \begin{array}{l} \begin{cases} -2 - m > 0 \\ \frac{m}{2} \notin [0, 1] \\ f(1) = \frac{m}{2 - m} = 2 \end{cases} \\ \begin{cases} -2 - m < 0 \\ \frac{m}{2} \notin [0, 1] \\ f(0) = \frac{1 + m}{-m} = 2 \end{cases} \end{array} \right.\)

\(\Leftrightarrow \left[ \begin{array}{l} \begin{cases} m < -2 \\ m \notin [0, 2] \\ m = \frac{4}{3} \end{cases} \\ \begin{cases} m > -2 \\ m \notin [0, 2] \\ m = -\frac{1}{3} \end{cases} \end{array} \right.\)\(\Leftrightarrow m = -\frac{1}{3}\)