In order to do this, we are going to start from the equation of the transistor
$I_E=I_B+I_C$
and since we know that $I_C/I_B=95$ so we can write

$$
frac{I_E}{I_B}=1+frac{I_C}{I_B}=1+95=boxed{96}
$$

$$
frac{I_E}{I_B}=96
$$

a) Since the voltage across the diode is constantly 0.7 V the entirety of the voltage increase will occur across the resistor, i.e. $boxed{textrm{1 V}}$.

b) Now, we can find that the current increase is given from the Ohm’s law as follows

$$
Delta I=frac{Delta V}{R}=boxed{frac{1textrm{ V}}{R}}
$$

a) The voltage across the diode is unchanged and across the resistor increases by 1 V

b) The increase in the current is $Delta I =frac{1 textrm{ V}}{R}$.

The $pn$-junction diodes by definition conduct currents much better when forward-biased i.e. they have much lower resistance in comparison when they are reverse-biased.

We can do so by using the definition of the current gain which is given as

$$
G=frac{I_C}{I_B}=frac{6.6times 10^{-3}}{0.055times 10^{-3}}=boxed{120}
$$

$$
G=120
$$

Although in principle, one can think of this as possible, in reality, due to the different characteristics of the electronic elements this is impossible. The reason for this is the fact that in diode the $p$-end is approximately of the same thickness as the $n$-end. So if we would merge two diodes together, the $p$-layer would be twice as thick as the respective $n$-ends. However, in an $npn$ transistor, $p$-end is thin enough to allow for the passage of electrons so the former construction would not resemble a transistor whatsoever.