Đáp án:
a) Ta có:
$\begin{array}{l}
f\left( {{x_1} + {x_2}} \right) = a\left( {{x_1} + {x_2}} \right) = a{x_1} + a.{x_2}\\
f\left( {{x_1}} \right) = a.{x_1}\\
f\left( {{x_2}} \right) = a.{x_2}\\
\Rightarrow f\left( {{x_1} + {x_2}} \right) = f\left( {{x_1}} \right) + f\left( {{x_2}} \right)\left( { = a.{x_1} + a.{x_2}} \right)
\end{array}$
b)
$\begin{array}{l}
f\left( {kx} \right) = a.kx = k.a.x\\
k.f\left( x \right) = k.\left( {a.x} \right) = k.a.x\\
\Rightarrow f\left( {kx} \right) = k.f\left( x \right)\left( {k \ne 0} \right)
\end{array}$