$\begin{array}{l}
1.\,\,{\sin ^2}A + {\sin ^2}B + {\sin ^2}C = \frac{{1 - c{\rm{os}}2A}}{2} + \frac{{1 - \cos
B}}{2} + 1 - c{\rm{o}}{{\rm{s}}^2}C\\
= 2 - \frac{{c{\rm{os}}2A + c{\rm{os}}2B}}{2} - c{\rm{o}}{{\rm{s}}^2}C = 2 -
c{\rm{os}}(A + B)c{\rm{os}}(A - B) - c{\rm{o}}{{\rm{s}}^2}C\\
= 2 + \cos C(c{\rm{os}}(A - B) - \cos C) = 2 + \cos C(c{\rm{os}}(A - B) + c{\rm{os}}(A + B))\\
= 2 + 2\cos C\cos A\cos B
\end{array}$
Nếu tam giác $ABC$ nhọn thì $cosA, cosB, cosC > 0$ $\Rightarrow\Delta
ABC$ nhọn
Vì vậy nếu tam giác ABC thỏa mãn điều kiện ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C > 2$ thì
tam giác $ABC$ nhọn.
$\begin{array}{l}
2.\,\,\left( {{\rm{1 }} + {\rm{ cosx}}} \right)\left( {{\rm{1 }} + {\rm{ sinx}}} \right){\rm{ }} = {\rm{ 2}}\\
\Leftrightarrow \cos x\sin x + \cos x + {\mathop{\rm s}\nolimits} {\rm{inx}} = 1(*)
\end{array}$
Đặt $t = cosx + sinx = \sqrt 2 c{\rm{os}}\left( {x - \frac{\pi }{4}} \right)$$ \Rightarrow |t| \le
\sqrt 2 $ và ${t^2} = 1 + 2\sin x\cos x$
$(*) \Leftrightarrow \frac{{{t^2} - 1}}{2} + t = 1$
$\Leftrightarrow
t^2+2t-3=0\Leftrightarrow\left[\begin{array}{I} t=1\\t=-3(L)\end{array}\right.$
$\Leftrightarrow\cos(x-\frac{\pi}{4})=\frac{1}{\sqrt{2}}$
$\Leftrightarrow \left[ \begin{array}{l}
x = 2k\pi \\
x = \frac{\pi }{2} + 2k\pi
\end{array} \right.\,\,\,\,k \in Z$