Đáp án:
$\begin{array}{l}
a)\overrightarrow {AD} .\overrightarrow {BC} + \overrightarrow {CA} .\overrightarrow {BE} + \overrightarrow {AB} .\overrightarrow {CF} \\
= \overrightarrow {BC} .\frac{1}{2}.\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right) + \overrightarrow {CA} .\frac{1}{2}.\left( {\overrightarrow {BC} + \overrightarrow {BA} } \right)\\
+ \overrightarrow {AB} .\frac{1}{2}.\left( {\overrightarrow {CB} + \overrightarrow {CA} } \right)\\
= \frac{1}{2}\left( \begin{array}{l}
.\overrightarrow {BC} .\overrightarrow {AB} + \overrightarrow {BC} .\overrightarrow {AC} + \overrightarrow {CA} .\overrightarrow {BC} + \\
\overrightarrow {CA} .\overrightarrow {BA} + \overrightarrow {AB} .\overrightarrow {CB} + \overrightarrow {AB} .\overrightarrow {CA}
\end{array} \right)\\
= \frac{1}{2}.\overrightarrow 0 = \overrightarrow 0 \\
b)\sin a + \cos a = \frac{1}{3}\\
\Rightarrow \sqrt 2 \sin \left( {a + \frac{\pi }{4}} \right) = \frac{1}{3}\\
\Rightarrow \sin \left( {a + \frac{\pi }{4}} \right) = \frac{1}{{3\sqrt 2 }}\\
\Rightarrow a = \arcsin \frac{1}{{3\sqrt 2 }} - \frac{\pi }{4}\\
\Rightarrow P = \sqrt {{{\tan }^2}a + {{\cot }^2}a} = \frac{{49}}{{16}}
\end{array}$
