Vtyjkyuk

I've tried this in Wolfram Alpha and got the value for $m$ that I expected, however I would like to know how to solve it rather than just the answer.

Note that $37$ is coprime to $phi(143)(=phi(11times13)=120$).
So in the multiplicative group of order 120, consisting integers coprime to $143, xmapsto x^37$ is an automorphism of groups (suffices to know it is a bijection).

Its inverse is the map $xmapsto x^a$ with $a$ being the inverse of 37 mod 120. (It happens that $a=13$ here.)

So, for $x$ coprime to 143, we can factorize the identity map ( Euler's theorem states $x^120$ is 1 mod 143) $xmapsto x ^1 = (x^13)^37pmod 143$

Applying for your case $x=108$, we get $m=108^13pmod143= 69pmod143$.

We work in the ring $R=Bbb Z/143$.
Its units form a group $R^times$ with $phi(143)=143cdotleft(1-frac 111right)left(1-frac 113right)=120$ elements. We want to solve in this group
$$
108=m^37 .
$$
We start from this, take the $13$.th power and get
$$
m=m^481=(m^37)^13=108^13=69
$$
in $R$. (We had to get the inverse of $37$ modulo $120$, it is $13$.)
One checks that this is a solution, to have the converse too.

sage search for the solution:

sage: R = Zmod(143)
sage: for m in R:
....: if m^37 == R(108):
....: print m
....: 
69
sage: R(108)^13
69

By Euler's theorem, $x^varphi (143)cong1pmod143$, if $x$ and $143$ are coprime.

$varphi (143)=varphi (13)cdot varphi (11)=(13-1)(11-1)=(12)(10)=120$.

Next, $(13)(37)-4(120)=1$. That is, $(13)(37)cong1pmod120$. (This can be done with the Euclidean algorithm, since $37$ and $120$ are coprime.)

Now $(x^37)^13=x^481=(x^120)^4cdot xcong xpmod143implies 108^13cong xpmod143implies x=69$. (For the last step I used Wolfram alpha.)

As $lambda(143)=60,$ let us find integer $a,b$ such that $$37a+60b=1$$

Like my answer here in Solving a Linear Congruence, $$37cdot13-60cdot8=1$$

$$m=m^37cdot13-60cdot8equiv(108)^13cdot1^-8pmod143$$

$108equiv-2pmod11implies(108)^13equiv(-2)^13equiv(-2)^3equiv3 (1)$ using Fermat's Little Theorem

Similarly, $(108)^13equiv4pmod13 (2)$

Apply Chinese Remainder Theorem, on $(1),(2)$