When the spring is cut into pieces, they will have the new force constant. The spring is divided into $$1:2:3$$   ratio.
When the pieces are connected in series, the resultant force constant
$$\eqalign{ & \frac{1}{{v'}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}} \cr & \frac{1}{{v'}} = \frac{1}{x} + \frac{1}{{2x}} + \frac{1}{{3x}} \cr & v' = \frac{{6x}}{{11}} \cr} $$
In parallel, the net force constant
$$K'' = x + 2x + 3x = 6x = 11K$$
The required ratio $$\frac{K}{{K''}} = \frac{{\frac{{6x}}{{11}}}}{{6x}} = 1:11$$

When $${F_1},{F_2}$$  and $${F_3}$$ are acting on a particle then the particle remains stationary. This means that the resultant of $${F_1},{F_2}$$  and $${F_3}$$ is zero, When $${F_1}$$ is removed, $${F_2}$$ and $${F_3}$$ will remain. But the resultant of $${F_2}$$ and $${F_3}$$ should be equal and opposite to $${F_1}.{\text{i}}{\text{.e}}{\text{.}}\left| {{{\vec F}_2} + {{\vec F}_3}} \right| = \left| {{{\vec F}_1}} \right|$$
$$\therefore a = \frac{{\left| {{{\vec F}_2} + {{\vec F}_3}} \right|}}{m} \Rightarrow a = \frac{{{F_1}}}{m}$$

Drawing free body-diagrams for $$m\& M,$$

we get $$T = ma$$   and $$F - T = Ma$$
where $$T$$ is force due to spring
$$\eqalign{ & \Rightarrow F - ma = Ma\,{\text{or,}}\,F = Ma + ma \cr & \therefore a = \frac{F}{{M + m}} \cr} $$
Now, force acting on the block of mass $$m$$ is
$$ma = m\left( {\frac{F}{{M + m}}} \right) = \frac{{mF}}{{m + M}}.$$

(i) When, mass is lifted upwards with an acceleration $$a,$$ then according to free body diagram

$$\eqalign{ & {T_1} - mg = ma \cr & \Rightarrow {T_1} = mg + ma \cr & {T_1} = m\left( {g + a} \right) \cr} $$
Substituting the values we obtain
$$\eqalign{ & \therefore {T_1} = \left( 1 \right)\left( {9.8 + 4.9} \right) \cr & = 14.7\,N \cr} $$
(ii) When, mass is lowered downwards with an acceleration $$a,$$ then

$$\eqalign{ & mg - {T_2} = ma \cr & \Rightarrow {T_2} = mg - ma = m\left( {g - a} \right) \cr} $$
Substituting the values, we have
$${T_2} = \left( 1 \right)\left( {9.8 - 4.9} \right) = 4.9\,N$$
Then, ratio of tensions $$\frac{{{T_1}}}{{{T_2}}} = \frac{{14.7}}{{4.9}} = \frac{3}{1}$$
$$ \Rightarrow {T_1}:{T_2} = 3:1$$

The tension in both strings will be same due to symmetry.

For equilibrium in vertical direction for body $$B$$ we have
$$\eqalign{ & \sqrt 2 \,mg = 2T\cos \theta \cr & \therefore \sqrt 2 \,mg = 2\left( {mg} \right)\cos \theta \,\left[ {\because T = mg,\,\left( {{\text{at}}\,{\text{equilibrium}}} \right.} \right] \cr & \therefore \cos \theta = \frac{1}{{\sqrt 2 }} \Rightarrow \theta = {45^ \circ } \cr} $$

Minimum force on $$A$$ = frictional force between the surfaces = $$12\,N$$
Therefore maximum acceleration
$${a_{\max }} = \frac{{12\,N}}{{4\,kg}} = 3\,m/{s^2}$$
Hence maximum force,
$${F_{\max }} = {\text{total}}\,{\text{mass}} \times {a_{\max }} = 9 \times 3 = 27\,N$$

The Earth pulls the block by a force $$Mg.$$  The block in turn exerts a force $$Mg$$  on the spring of spring balance $${S_1}$$ which therefore shows a reading of $$M\,kgf.$$
The spring $${S_1}$$ is massless. Therefore it exerts a force of $$Mg$$  on the spring of spring balance $${S_2}$$ which shows the reading of $$M\,kgf.$$

$$\eqalign{ & \vec p\left( t \right) = A\left[ {\hat i\cos \left( {kt} \right) - \hat j\sin \left( {kt} \right)} \right] \cr & \vec F = \frac{{d\vec p}}{{dt}} = Ak\left[ { - \hat i\sin \left( {kt} \right) - \hat j\cos \left( {kt} \right)} \right] \cr & {\text{Here,}}\,\vec F.\vec P = 0\,\,{\text{But}}\,\,\vec F.\vec P = Fp\cos \theta \cr & \therefore \cos \theta = 0 \Rightarrow \theta = {90^ \circ } \cr} $$

$$\eqalign{ & F = ma \cr & \Rightarrow a = \frac{F}{m} = \frac{{5 \times {{10}^4}}}{{3 \times {{10}^7}}} = \frac{5}{3} \times {10^{ - 3}}m{s^{ - 2}} \cr & {\text{Also, }}{v^2} - {u^2} = 2as \cr & \Rightarrow {v^2} - {0^2} = 2 \times \frac{5}{3} \times {10^{ - 3}} \times 3 = {10^{ - 2}} \cr & \Rightarrow v = 0.1m{s^{ - 1}} \cr} $$

Tension in both string shall be same which can be observed by making $$FBD$$  of string in figure (i)