Vtyjkyuk

$$largefracmu b2frac2-nu1-nuRlnfracRr_c-taupi R^2+fracgamma_0pi^2 R^2b_0sinleft(frac2pi(u_0+b)b_0right)=0$$

$$largefracmu b^24frac2-nu1-nuleft[1+lnfracRr_cright]-2pi Rtau b-gamma_0pi Rsinleft(fracpi(2u_0+b)b_0right)sinleft(fracpi bb_0right)=0$$

I've tried inputting the two equations in Mathematica but I think my input formatting is incorrect.

Using Mefitico's answer, extract $tau$ from the first equation and plug the resulting expression in the second equation which is now on the form $f(b)=0$. Solve it for $b$ (I suppose that graphing the function would be a good idea to locate more or less where is the root) and use any numerical method of your choice for solving it.

No hope for analytical solutions.

First, you re-write your equations to have coefficients $c_k$ instead of large expressions:

$$
c_1 b + c_2 tau+c_3 sin(c_4 +c_5 b) = 0
$$

$$
c_6 b^2 + c_7 b tau+c_8 sin(c_9 +c_10 b) sin(c_11b) = 0
$$

Now this should be much easier to write on Mathematica, and you can define each symbol and check for its appropriate typing.

Then, since the first equation looks a bit like Kepler's Equation, I would not hope to find an analytical solution, but depending on your coefficients you should be able to find a numeric solution with rather simple methods such as fixed point iteration or in the worst case Levenberg-Marquardt.