Answer
$79.3^\circ, 14.7^\circ$
Work Step by Step
As we see in the figures below, the two balls will be released from the same height $y_i=h$ since they have moved at the same angle and they have the same string length.
At the lowest point, at the bottom of their circular path, the momentum is conserved just before and just after the collision.
They met heads on in the horizontal direction.
$$p_{ix}=p_{fx}$$
$$m_1v_{ix,1}+m_2v_{ix,2}=m_1v_{fx,1}+m_2v_{fx,2} $$
Plugging the known;
$$0.1v_{ix,1}+0.2v_{ix,2}=0.1v_{fx,1}+0.2v_{fx,2} $$
Multiplying both sides by 10;
$$ v_{ix,1}+2v_{ix,2}= v_{fx,1}+2v_{fx,2} \tag 1$$
We can use energy conservation to find their velocities just before they hit each other.
$$E_{i1}=E_{f1}$$
$$K_{i1}+U_{gi,1}=K_{f1}+U_{gf,1}$$
we know that the two balls are released from rest, so their initial kinetic energy is zero. We also know that they met each other at the bottom which we chose to be the origin at which $y=0$, so their final gravitational potential energy is zero.
$$0+U_{gi,1}=K_{f1}+0$$
$$\color{red}{\bf\not}m_1gy_{i1} =\frac{1}{2}\color{red}{\bf\not}m_1v_{fx,1}^2 $$
From the figure below, we can see that $y_{i1}=y_{i2}=h$
$$ gh=\frac{1}{2} v_{fx,1}^2 $$
where $h$ is given by $\cos45^\circ =\dfrac{L-h}{L}$, and hence, $L\cos45^\circ=L-h\Rightarrow \boxed{h=L(1-\cos45^\circ)}$
$$ gL(1-\cos45^\circ)=\frac{1}{2} v_{fx,1}^2 $$
Solving for $v_{fx,1}$;
$$v_{fx,1}=\sqrt{ 2gL(1-\cos45^\circ) }$$
Noting that since the two balls are released from the same height and since we can see that the velocity of the ball has no relation to its mass, so both balls must have the same speed just before the collision.
Thus,
$$v_{fx,1}=v_{fx,2}=\sqrt{ 2gL(1-\cos45^\circ) }$$
Plug the known;
$$v_{1}=v_{2}=\sqrt{ 2(9.8)(1)(1-\cos45^\circ) }=\bf 2.396 \;\rm m/s$$
Plugging these two speeds, which are the speeds of the two balls just before the collision, into (1). Note the signs, the ball that moves to the right must has a positive velocity while the other has a negative velocity.
$$ 2.396 +2(-2.396 )= v_{fx,1}+2v_{fx,2} $$
$$ -2.396 = v_{fx,1}+2v_{fx,2} \tag 2$$
We assume that the collision is perfectly elastic, so the energy before and after the collision is conserved as well.
$$E_i=E_f$$
$$K_{i}+U_{ig}=K_{f}+U_{fg}$$
The tow balls are at the bottom before and after the collision, so their gravitational potential energies are zeros.
$$K_{i}+0=K_{f}+0$$
$$\color{red}{\bf\not}\frac{1}{2} m_1v_{ix,1}^2+\color{red}{\bf\not}\frac{1}{2} m_2v_{ix,2}^2=\color{red}{\bf\not}\frac{1}{2} m_1v_{fx,1}^2+\color{red}{\bf\not}\frac{1}{2} m_2v_{fx,2}^2 $$
We know that $v_{ix,1}=v_{ix,2}=v_{ix}=2.396$ m/s.
$$ (m_1 + m_2)v_{ix}^2= m_1v_{fx,1}^2+ m_2v_{fx,2}^2 $$
Plugging the known;
$$ (0.1+ 0.2) 2.396^2= 0.1v_{fx,1}^2+ 0.2v_{fx,2}^2 $$
$$1.722= 0.1v_{fx,1}^2+ 0.2v_{fx,2}^2 $$
Plugging $v_{fx,1}$ from (2);
$$1.722= 0.1\left[ -2.396 -2v_{fx,2} \right]^2+ 0.2v_{fx,2}^2 $$
$$1.722= 0.1\left[ 2.396^2+ 4.792v_{fx,2} +4v_{fx,2}^2 \right] + 0.2v_{fx,2}^2 $$
$\times 10$ for both sides;
$$17.22= 2.396^2+ 4.792v_{fx,2} +4v_{fx,2}^2 + 2v_{fx,2}^2 $$
Thus,
$$0= 6v_{fx,2}^2 + 9.584 v_{fx,2} +2.396^2-17.22$$
Hence,
$$v_{fx,2}=-2.39587\;\rm m/s, \;Or\;0.798539\;m/s$$
We know that the second ball after the collision will move to the right which is the positive direction, so we can neglect the negative root here.
$$v_{fx,2}= \bf 0.7985\rm\;m/s\tag 3$$
And hence, plugging into (2);
$$v_{fx,1}= \bf -3.994\rm\;m/s\tag 4$$
Now we can chose the first ball as an isolated system after the collision where its energy is conserved.
$$E_{i1}=E_{f1}$$
$$K_{i1}+U_{ig,1}=K_{f1}+U_{fg,1}$$
The ball will move in vertical circular path until it stops, so its final kinetic, energy when it makes an angle of $\theta_1$ with the vertical line, is zero. And its starts from the bottom, so its initial kinetic energy is zero.
$$K_{i1}+0=0+U_{fg,1}$$
$$\frac{1}{2}\color{red}{\bf\not}m_1v_{fx,1}^2 =\color{red}{\bf\not}m_1gh_1$$
where $h_1=L(1-\cos\theta_1)$
$$\frac{1}{2} v_{fx,1}^2 = gL(1-\cos\theta_1) $$
Thus,
$$ \cos\theta_1=1-\dfrac{v_{fx,1}^2 }{2gL} $$
$$\theta_1= \cos^{-1}\left(1-\dfrac{v_{fx,1}^2 }{2gL} \right)\tag 5$$
Plugging from (4) and plug the other known;
$$\theta_1= \cos^{-1}\left(1-\dfrac{ (-3.994)^2 }{2(9.8)(1)} \right)=\color{red}{\bf 79.3}^\circ$$
By the same approach, as (5), for the second ball,
$$\theta_2= \cos^{-1}\left(1-\dfrac{v_{fx,2}^2 }{2gL} \right) $$
Plugging from (3) and plug the other known;
$$\theta_2= \cos^{-1}\left(1-\dfrac{ 0.7985^2 }{2(9.8)(1)} \right)=\color{red}{\bf 14.7}^\circ$$
