Partial answer ...
Number System 1:
This is basically a base 95 number system where the value of each digit is the corresponding ASCII value minus 32.
$9 => (57-32)*95^0=25$
$\\\$, => (36-32)*95^1+(44-32)*95^0=392$
$W+ => (87-32)*95^1+(43-32)*95^0=5236$
$00t => (48-32)*95^2+(48-32)*95^1+(116-32)*95^0=146004$
$ >3V1 => (62-32)*95^3+(51-32)*95^2+(86-32)*95^1+(49-32)*95^0=25897872$
$$\small\begin{array}{rrrrr}\\ &9 \rightarrow &&&&(57-32) \cdot 95^0=&25\\&\\\$, \rightarrow &&&(36-32) \cdot 95^1 + &(44-32) \cdot 95^0=&392\\&W+ \rightarrow &&&(87-32) \cdot 95^1 + &(43-32) \cdot 95^0=&523600\\&t \rightarrow &&(48-32) \cdot 95^2 + &(48-32) \cdot 95^1 + &(116-32) \cdot 95^0=&146004\\&>3V1 \rightarrow &(62-32) \cdot 95^3 + &(51-32) \cdot 95^2 + &(86-32) \cdot 95^1 + &(49-32) \cdot 95^0=&25897872\\\end{array}$$
Number System 2:
In this system the digits have the following values from least to most significant:
In this system the digits have the following values from least to most significant:
$1^0 \space 2^1 \space 3^2 \space 4^3 \space ...$
Interpreting $C$ as$1^0 \space 2^1 \space 3^2 \space 4^3 \space ...$
Interpreting $C$ as $12$ similar to hexadecimal numbers this gives:
$12$ similar to hexadecimal numbers this gives:
$2*3^2+3*2^1+1*1^0=25$
$6*4^3+0*3^2+4*2^1+0*1^0=392$
$8*5^4+3*4^3+4*3^2+4*2^1+0*1^0=5236$
$1*7^6+3*6^5+8*5^4+0*4^3+3*3^2+0*2^1+0*1^0=146004$
$12*8^7+6*7^6+3*6^5+4*5^4+5*4^3+0*3^2+3*2^1+0*1^0=25897872$$$\small\begin{array}{rrrrrrrrrr}\\&&&&&&2 \cdot 3^2 + &3 \cdot 2^1 + &1 \cdot 1^0 = &25\\&&&&&6 \cdot 4^3 + &0 \cdot 3^2 + &4 \cdot 2^1 + &0 \cdot 1^0 = &392\\&&&&8 \cdot 5^4 + &3 \cdot 4^3 + &4 \cdot 3^2 + &4 \cdot 2^1 + &0 \cdot 1^0 = &5236\\&&1 \cdot 7^6 + &3 \cdot 6^5 + &8 \cdot 5^4 + &0 \cdot 4^3 + &3 \cdot 3^2 + &0 \cdot 2^1 + &0 \cdot 1^0 = &146004\\&12 \cdot 8^7 + &6 \cdot 7^6 + &3 \cdot 6^5 + &4 \cdot 5^4 + &5 \cdot 4^3 + &0 \cdot 3^2 + &3 \cdot 2^1 + &0 \cdot 1^0 = &25897872\end{array}$$
