(a) If $i$ is the current in the diode and $V$ is voltage drop across it, then for given figure voltage equation is
$i \times 100 + V = 8 \Rightarrow i = - \frac{1}{100}V + \frac{8}{100}$
$\Rightarrow i = - (0.01)V + 0.08$
Thus the slope of $i$-$V$ graph $= \frac{1}{R_{L}} = 0.01$