For same range angles of projection should be $\theta$ and 90 - $\theta$
So, time of flights $t_1$ = $\frac{2usin\theta }{g}$ and
$t_2$ = $\frac{2usin(90-\theta )}{g}$ = $\frac{2ucos\theta }{g}$
By multiplying = $t_1{t}_2$ = $\frac{4u^{2}sin\theta cos\theta }{g^2}$
$t_1{t}_2$  = $\frac{2}{g}$ $\frac{\left(u^{2}sin2\theta \right)}{g}$ = $\frac{2R}{g}$ $\Rightarrow$ $t_1{t}_2$ $\alpha$ R

Instantaeous speed of rising mass after t sec will be $v_t$ = $\sqrt{{v_x}^2+{v_y}^2}$
where $v_x$ = v sin $\theta$ = Horizontal component of velocity 
$v_y$ = v sin $\theta$ - gt = Vertical component of velocity
$v_t$ = $\sqrt{{\left(vcos\theta \right)}^2+{(vsin\theta -gt)}^2}$

T sin $\theta$ = M ${\omega }^2$ R
T sin $\theta$ = M ${\omega }^2$ L sin $\theta$
From (i) and (ii)
T = M ${\omega }^2$ L
= M 4${\pi }^2{n}^{2}L$
= M 4 ${\pi }^2\left({\frac{2}{\pi }}^2\right)L$
= 16 ML
 

Since the maximum tension $T_B$ in the string moving in the vertical circle is at the bottom and minimum tension $T_T$ is at the top.
$T_B$ = $\frac{m{v_B}^2}{L}$ = mg  and $T_T$ = $\frac{m{v_T}^2}{L}$ - mg
$\therefore$ $\frac{T_B}{T_T}$ = $\frac{\frac{m{v_B}^2}{L}+mg}{\frac{m{v_T}^2}{L}-mg}$ or $\frac{{v_B}^2+gL}{{v_T}^2+gL}$ = $\frac{4}{1}$
or ${v_B}^2$ + gL = 4${v_T}^2$ - 4gL but ${v_B}^2$ = ${v_T}^2$ + 4gL
$\therefore$ ${v_T}^2$ + 4gL +gL $\Rightarrow$ 3 ${v_T}^2$ = 3gL
$\therefore$ ${v_T}^2$ = 3 $\times$ g $\times$ L = 3 $\times$ 10 $\times$ $\frac{10}{3}$ or $v_T$ = 10 m/sec

Let ${\hat{n}}_1$ and ${\hat{n}}_2$ are the two unit vectors, then the sum is
${\vec{n}}_s$ = ${\hat{n}}_1$ + ${\hat{n}}_2$ or ${n_s}^2$ = ${n_1}^2$ + ${n_2}^2$ + 2$n_1{n}_{2}cos\theta$
= 1 + 1 + 2cos$\theta$
Since it is given that $n_s$ is also a unit vector, therefore 1 = 1+ 1 + 2 cos $\theta$ = $\Rightarrow$ cos $\theta$ = - $\frac{1}{2}$ $\therefore$ $\theta$ = ${120}^{\circ }$
Now the difference vector is ${\hat{n}}_d$ = ${\hat{n}}_1$ - ${\hat{n}}_2$ or ${n_1}^2$ + ${n_2}^2$ - 2$n_1{n}_{2}cos\theta$ = 1 + 1 - 2 cos(${120}^{\circ }$) 
$\therefore$ ${n_d}^2$ = 2 - 2(- $\frac{1}{2}$) = 2 + 1 = 3 $\Rightarrow$ $n_d$ = $\sqrt{3}$

$\frac{B}{2}$ = $\sqrt{A^2+B^2+2ABcos\theta }$ ..(i)
$\therefore$ tan ${90}^{\circ }$ = $\frac{Bsin\theta }{A+Bcos\theta }$ $\Rightarrow$ A + Bcos$\theta$ = 0
$\therefore$ cos $\theta$ = - $\frac{A}{B}$
Hence, from (i) $\frac{B^2}{4}$ = $A^2+B^2-2A^2$ $\Rightarrow$ A = $\sqrt{3}\frac{B}{2}$
$\Rightarrow$ cos $\theta$ = - $\frac{A}{B}$ = - $\frac{\sqrt{3}}{2}$ $\therefore$ $\theta$ = ${150}^{\circ }$

If the magnitude of vector remains same, only direction change by $\theta$ then
$\overrightarrow{△}v$ = ${\vec{v}}_2$ - ${\vec{v}}_1$, $\overrightarrow{△}v$ = ${\vec{v}}_2$ + (- ${\vec{v}}_1$)
Magnitude of change in vector |$\overrightarrow{△}v$| = 2 v sin ($\frac{\theta }{2}$)
|$\overrightarrow{△}v$| = 2 $\times$ 10 $\times$ sin($\frac{{90}^{\circ }}{2}$) = 10 $\sqrt{2}$ = 14.14 m/s
Direction is south - west as shown in figure.

$R_{max}$ = A + B = 17 when $\theta$ = $0^{\circ }$
$R_{min}$ = A - B = 7 when $\theta$ = ${180}^{\circ }$
by solving we get A = 12 and B = 5
Now when $\theta$ = ${90}^{\circ }$ then R = $\sqrt{A^2+B^2}$
$\Rightarrow$R=$\sqrt{{\left(12\right)}^2+{\left(5\right)}^2}$ = $\sqrt{169}$ = 13

Let P be the smaller force and Q be the greater force then according to problem-
P + Q = 18 ..(i)
R = $\sqrt{P^2+Q^2+2PQcos\theta }$ = 12 ..(ii)
tan $\varphi$ = $\frac{Qsin\theta }{P+Qcos\theta }$ = tan 90 = $\infty$
$\therefore$ P + Q cos $\theta$ = 0 ..(iii)
By solving (i), (ii) and (iii) we will get P = 5 , and Q = 13

As the metal sphere is in equilibrium under the effect of three forces therefore $\overrightarrow{T}$ + $\overrightarrow{P}$ + $\overrightarrow{W}$ = 0
From the figure T cos $\theta$ =W ...(i)
T sin $\theta$ = P ..(ii)
From equation (i) and (ii) we get P= Wtan $\theta$
and $T^2$ = $P^2$ + $W^2$