Thanks for providing all your work effort. It's not as commonly done as many of us would wish.

You are able to use the schematic editor to produce a more readable version and you should learn how to do that as soon as you can, if you plan to ask more questions here. And you can use MathJax to write out your equations, as well. (As I have to, in fact.)

To keep my time down, I'm not creating a schematic for you. It would be nice, but I'm short on time. I'll use \$I_1\$ for the current in loop 1, etc. (Please do note that this means I will not be following all of your designations on your paper as you have, for example, an arrow with \$I_1\$ labeled on it that has nothing to do with the current I will label \$I_1\$. Just be aware.) And I'll also start at the tail end of your loop arrows and work myself around in the indicated directions, as well. Just so you can follow along with my reasoning:

$$\begin{align*} 0\:\text{V} + 9\:\text{V} - \left(I_1+I_2\right)\cdot 90\:\Omega - I_1\cdot 60\:\Omega-I_1\cdot 30\:\Omega&=0\:\text{V}\\\\ 0\:\text{V}+12\:\text{V}-\left(I_2-I_3\right)\cdot 120\:\Omega-\left(I_1+I_2\right)\cdot 90\:\Omega&=0\:\text{V}\\\\ 0\:\text{V} + 9\:\text{V} - \left(I_3-I_2\right)\cdot 120\:\Omega-I_3\cdot 30\:\Omega&=0\:\text{V} \end{align*}$$

You should be able to compare the results of solving those three equations with what you wrote down and solved. (I can't read your blurry figures towards the end of your photo, sorry.) From the above currents, you can add and subtract them as required and work out various voltage drops across the resistors and from there pretty much solve anything of interest in the schematic.

Thanks for providing all your work effort. It's not as commonly done as many of us would wish.

You are able to use the schematic editor to produce a more readable version and you should learn how to do that as soon as you can, if you plan to ask more questions here. And you can use MathJax to write out your equations, as well. (As I have to, in fact.)

To keep my time down, I'm not creating a schematic for you. It would be nice, but I'm short on time. I'll use \$I_1\$ for the current in loop 1, etc. (Please do note that this means I will not be following all of your designations on your paper as you have, for example, an arrow with \$I_1\$ labeled on it that has nothing to do with the current I will label \$I_1\$. Just be aware.) And I'll also start at the tail end of your loop arrows and work myself around in the indicated directions, as well. Just so you can follow along with my reasoning:

$$\begin{align*} 0\:\text{V} + 9\:\text{V} - \left(I_1+I_2\right)\cdot 90\:\Omega - I_1\cdot 60\:\Omega-I_1\cdot 30\:\Omega&=0\:\text{V}\\\\ 0\:\text{V}+12\:\text{V}-\left(I_2-I_3\right)\cdot 120\:\Omega-\left(I_1+I_2\right)\cdot 90\:\Omega&=0\:\text{V}\\\\ 0\:\text{V} + 9\:\text{V} - \left(I_3-I_2\right)\cdot 120\:\Omega-I_3\cdot 30\:\Omega&=0\:\text{V} \end{align*}$$

You should be able to compare the results of solving those three equations with what you wrote down and solved. (I can't read your blurry figures towards the end of your photo, sorry.) From the above currents, you can add and subtract them as required and work out various voltage drops across the resistors and from there pretty much solve anything of interest in the schematic.

Just added this since you added more work to your question.

The solution to the above is (I'm using sympy):

var('i1 i2 i3')
eq1=Eq(0+9-(i1+i2)*90-i1*60-i1*30,0)
eq2=Eq(0+12-(i2-i3)*120-(i1+i2)*90,0)
eq3=Eq(0+9-(i3-i2)*120-i3*30,0)
solve([eq1,eq2,eq3],[i1,i2,i3])
{i1: -13/230, i2: 49/230, i3: 53/230}

The above checks out with a Spice simulation, perfectly.