For mass $${m_1}$$
$${m_1}g - T = {m_1}a$$
For mass $${m_2}$$
$$T - {m_2}g = {m_2}a$$

Adding the equations we get
$$\eqalign{ & a = \frac{{\left( {{m_1} - {m_2}} \right)g}}{{{m_1} + {m_2}}} \cr & \therefore \frac{1}{8} = \frac{{\frac{{{m_1}}}{{{m_2}}} - 1}}{{\frac{{{m_1}}}{{{m_2}}} + 1}} \Rightarrow \frac{{{m_1}}}{{{m_2}}} + 1 = 8\frac{{{m_1}}}{{{m_2}}} - 8 \Rightarrow \frac{{{m_1}}}{{{m_2}}} = \frac{9}{7} \cr} $$

Assuming both the blocks are stationary
$$N = F$$
$$\eqalign{ & {f_1} = 20\,{\text{N}} \cr & {f_1} = 100 + 20 = 120\,N \cr} $$

Considering the two blocks as one system and due to equilibrium $$f = 120 N$$

$$\eqalign{ & {v^2} - {u^2} = 2as\,{\text{or}}\,{0^2} - {u^2} = 2\left( { - {\mu _k}g} \right)s \cr & - {100^2} = 2 \times - \frac{1}{2} \times 10 \times s \Rightarrow s = 1000m \cr} $$

In situation 1, the tension $$T$$ has to hold both the masses $$2m$$ and $$m$$ therefore,
$$T = 3mg$$
In situation 2, when the string is cut, the mass $$m$$ is a freely falling body and its acceleration due to gravity is $$g.$$
For mass $$2m,$$  just after the string is cut, $$T$$ remains $$3mg$$  because of the extension of string.
$$\therefore 3mg - 2mg = 2m \times a\,\,\,\,\therefore \frac{g}{2} = a$$

Let $$T$$ be the tension in the string.
$$\eqalign{ & \therefore 10g - T = 10a\,......\left( {\text{i}} \right) \cr & T - 5g = 5a\,......\left( {{\text{ii}}} \right) \cr} $$
Adding (i) and (ii),
$$5g = 15a \Rightarrow a = \frac{g}{3}m/{s^2}$$

$$\eqalign{ & ma\cos {30^ \circ } = mg\sin {30^ \circ } \cr & \therefore a = \frac{g}{{\sqrt 3 }} \cr} $$