Solutions:

Electrons are 'boiled off'a hot metal plate because of the Edison-Richardson-Effect and made up an 'electron gas' around the filament. The heater voltage determines the temperature of the filament. The higher the heater voltage the higher the temperature of the filament and the more electrons are 'boiled off' the filament. So more electrons can flow from the hot cathode to the anode. The electron beam is getting more intense when the heater voltage rises.

The acceleration voltage determines the electric field between hot cathode and anode. So the acceleration voltage causes the acceleration of the electrons in the direction of the anode. The higher the acceleration voltage the faster the electrons get when passing the anode.

Solutions:

1. $${v=\sqrt{2\cdot \frac{e}{m}\cdot V_{\text a}}\Rightarrow v=\sqrt{2\cdot \frac{1.6\cdot 10^{-19}\,\text{C}}{9.1\cdot 10^{-31}\,\text{kg}}\cdot 500 \,\text{V}}\approx 1.326\cdot 10^7\,\frac{\text m}{\text s}}$$
2. $$W_{\text{el}}=F_{\text{el}}\cdot \text {d} = V_{\text a}\cdot e \Rightarrow W_{\text{el}}=500 \,\text V \cdot 1.6\cdot 10^{-19}\,\text C \approx 8.0 \cdot 10^{-17} \,\text J$$
3. For uniformly accelerated motions is: $$s=\frac{1}{2} a\cdot t^2 \Rightarrow t=\sqrt{\frac{2s}{a}}$$ The force acting on electrons in an electric field is $$F_{\text{el}}=\frac{V_{\text a}\cdot e}{\text d}.$$ With $F=m\cdot a$ the acceleration of the elctrons is: $$a=\frac{F_{\text{el}}}{m_{\text e}}=\frac{V_{\text a}\cdot e}{{m_{\text e}\cdot \text d}}$$ Putting this in our starting formula we find: $$t=\sqrt{\frac{2\cdot s}{\frac{V_{\text a}\cdot e}{{m_{\text e}\cdot \text d}}}}=\sqrt{\frac{2\cdot s\cdot m_{\text e}\cdot \text d}{V_{\text a}\cdot e}}\Rightarrow t=\sqrt{\frac{2\cdot 9.1\cdot 10^{-31}\,\text{kg}\cdot 0.05\,\text m\cdot 0.05\,\text m}{500\,\text V\cdot 1.6\cdot 10^{-19}\text C}} \approx 7.54\cdot 10^{-9}\,\text s$$
4. $${v=\sqrt{2\cdot \frac{e}{m}\cdot V_{\text a}}\Leftrightarrow V_{\text a}=\frac{v^2\cdot m}{2\cdot e} \Rightarrow V_{\text a}=\frac{ \left( { 2.054\cdot 10^7\,\frac{\text m}{\text s}} \right)^2 \cdot 9.1\cdot 10^{-31}\,\text{kg}}{2\cdot 1.6\cdot 10^{-19}\,\text C}\approx 1200\,\text V}$$

Solutions:

1. A relativistic calculation must be done because the acceleration voltage is higher than 2.7 kV and so the spped of the electrons is higher than 10% of the speed of light. So non-relativistic calculation lead to a non negligibly error.
   
2. $${v_{\text{rel}}=c\cdot \sqrt{1-\frac{1}{\left( {1+\frac{V_{\text a}\cdot e}{m_{e}\cdot c^2}}\right)^2}}}$$ $$\Rightarrow v_{\text{rel}}=3\cdot 10^{8}\,\frac{\text m}{\text s}\sqrt{1-\frac{1}{ \left( 1+\frac{8000\,\text{V}\cdot 1.6\cdot 10^{-19}\,\text{C}}{9.1\cdot 10^{-31}\,\text{kg}\cdot \left( 3\cdot 10^8\,\frac{\text m}{\text s}\right)^2 }\right)^2}}\approx 0.1745\cdot c = 5.243\cdot 10^7\,\frac{\text m}{\text s} $$
3. $$ V_{\text a}=\left( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right)\cdot\frac{m_e\cdot c^2}{e}\Rightarrow V_{\text a}\approx 79.187\,\text {kV}$$
4. non-relativistic: $$V_{\text a}=\frac{v^2\cdot m}{2\cdot e} \Rightarrow V_{\text a}=\frac{c^2\cdot m}{8\cdot e}\approx 63.984\,\text {kV}$$ relative error \(f\): $$f=\frac{V_{\rm{class}}}{V_{\rm{rel}}}-1\Rightarrow\frac{63984\,\text V}{79187\,\text V}-1\approx -0.192=-19.2\,\%$$With non-relativistic calculation the acceleration voltage would be about \(19.2\%\) too low.