I may not be true. Let $a=\pi$, which is a real number. Then we have
$\begin{array}{rcl}
uv & = & \dfrac{7}{\pi+i} \times \dfrac{7}{\pi – i} \\
& = & \dfrac{49}{\pi^2 -i^2} \\
& = & \dfrac{49}{\pi^2 +1} \ \text{, which is a irrational number.}
\end{array}$
II must be true. For any real number $a$,
$\begin{array}{rcl}
u & = & \dfrac{7}{a+i} \\
& = & \dfrac{7}{a+i} \times \dfrac{a-i}{a-i} \\
& = & \dfrac{7a – 7i}{a^2 -i^2} \\
& = & \dfrac{7a}{a^2 +1} -\dfrac{7}{a^2+1}i
\end{array}$
$\begin{array}{rcl}
v & = & \dfrac{7}{a-i} \\
& = & \dfrac{7}{a-i} \times \dfrac{a+i}{a+i} \\
& = & \dfrac{7a + 7i}{a^2 -i^2} \\
& = & \dfrac{7a}{a^2 +1} +\dfrac{7}{a^2+1}i
\end{array}$
Hence, the real part of $u$ is equal to the real part of $v$.
III may not be true. Let $a=0$, which is a real number.
$\begin{array}{rcl}
\dfrac{1}{u} & = & \dfrac{1}{7} i \\
\dfrac{1}{v} & = & \dfrac{-1}{7} i
\end{array}$
Hence, the imaginary part of $\dfrac{1}{u}$ is not equal to the imaginary part of $\dfrac{1}{v}$.
