$$\large{\frac{\mu b}2\frac{2-\nu}{1-\nu}R\ln\frac{R}{r_c}-\tau\pi R^2+\frac{\gamma_0\pi^2 R^2}{b_0}\sin\left(\frac{2\pi(u_0+b)}{b_0}\right)=0}$$
$$\large{\frac{\mu b^2}4\frac{2-\nu}{1-\nu}\left[1+\ln\frac{R}{r_c}\right]-2\pi R\tau b-\gamma_0\pi R\sin\left(\frac{\pi(2u_0+b)}{b_0}\right)\sin\left(\frac{\pi b}{b_0}\right)=0}$$
I've tried inputting the two equations in Mathematica but I think my input formatting is incorrect.
On
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:
$$ c_1 b + c_2 \tau+c_3 \sin(c_4 +c_5 b) = 0 $$
$$ c_6 b^2 + c_7 b \tau+c_8 \sin(c_9 +c_{10} b) \sin(c_{11}b) = 0 $$
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.