Đáp án:

$\begin{array}{l}
a)Dkxd:a - 1 \ge 0 \Leftrightarrow a \ge 1\\
Vậy\,a \ge 1\\
b)DKxd:\left( {a - 3} \right)\left( {a + 5} \right) \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a - 3 \ge 0\\
a + 5 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
a - 3 \le 0\\
a + 5 \le 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
a \ge 3\\
a \le  - 5
\end{array} \right.\\
Vậy\,a \le  - 5\,hoac\,a \ge 3\\
c)Dkxd:1 - 3a \ge 0\\
 \Leftrightarrow 3a \le 1\\
 \Leftrightarrow a \le \dfrac{1}{3}\\
Vậy\,a \le \dfrac{1}{3}\\
d)Dkxd:\left\{ \begin{array}{l}
a \ge 0\\
1 - 2\sqrt a  + a \ge 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
a \ge 0\\
{\left( {1 - \sqrt a } \right)^2} \ge 0
\end{array} \right.\\
Vậy\,a \ge 0\\
e)\dfrac{3}{{a + 5}} \ge 0\\
 \Leftrightarrow a + 5 > 0\\
 \Leftrightarrow a >  - 5\\
Vậy\,a >  - 5\\
f)\dfrac{{a - 3}}{{2a - 5}} \ge 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a - 3 \ge 0\\
2a - 5 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
a - 3 \le 0\\
2a - 5 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a \ge 3\\
a > \dfrac{5}{2}
\end{array} \right.\\
\left\{ \begin{array}{l}
a \le 3\\
a < \dfrac{5}{2}
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
a \ge 3\\
a < \dfrac{5}{2}
\end{array} \right.\\
Vậy\,a < \dfrac{5}{2}\,hoac\,a \ge 3
\end{array}$