I would simplify your code this way, if I have guessed right what you meant:

Clear["Global`*"]
eqn1 = \[Tau]*p''[x] - a*p[x] + \[Rho]*\[Mu]'[x] == 0;
eqn2 = \[Mu]''[x] + \[Delta]*
     p'[x] + \[Nu]*\[Mu][x] - \[Nu]*\[Mu]hat == 0;
{p, \[Mu]} = DSolveValue[{eqn1, eqn2}, {p, \[Mu]}, x];
u = DSolveValue[
   u'[x] - (1/\[Zeta])*(\[Sigma] - 
        h*p'[x] + (\[Alpha]/\[Beta])*\[Mu][
          x] - (\[Alpha]/\[Beta])*\[Mu]hat) == 0, u, x];
eqn3 = h*u'[(w/2)] + b*p'[(w/2)] == 0;
eqn4 = h*u'[-(w/2)] + b*p'[-(w/2)] == 0;
eqn5 = u[0] == 0;
eqn6 = \[Mu][(w/2)] == \[Mu]1;
eqn7 = \[Mu][-(w/2)] == \[Mu]0;
Solve[{eqn3, eqn4, eqn5, eqn6, eqn7}, {C[5], C[4], C[3], C[2], C[1]}]

I didn't wait for the last equation to give an answer, it took too long. The full system is linear in the unknowns, but it has very complicated coefficients.

I only tried with some numerical values for the parameters:

values = {\[Tau] -> 1, 
   a -> 1, \[Rho] -> 2, \[Delta] -> 1, \[Nu] -> 2, \[Mu]hat -> 
    1, \[Zeta] -> 1, \[Sigma] -> 1, 
   h -> 1, \[Alpha] -> 1, \[Beta] -> 1, b -> 1, 
   w -> 1, \[Mu]0 -> \[Mu]0, \[Mu]1 -> \[Mu]1};
eqs = Collect[{eqn3, eqn4, eqn5, eqn6, eqn7} /. values, C[_], Simplify]
{knowns, mat} = CoefficientArrays[eqs, Array[C, 5]]
Det[mat]
eqs // N // Simplify // Chop
Solve[eqs, Array[C, 5]]

In this case the coefficient matrix has zero determinant. No solution exists.

For other values of the parameters the derterminant is nonzero and a solution exists:

values = {\[Tau] -> 1, 
   a -> 1, \[Rho] -> 2, \[Delta] -> 1, \[Nu] -> 1, \[Mu]hat -> 
    1, \[Zeta] -> 1, \[Sigma] -> 1, 
   h -> 1, \[Alpha] -> 1, \[Beta] -> 1, b -> -1, 
   w -> 3, \[Mu]0 -> \[Mu]0, \[Mu]1 -> \[Mu]1};
eqs = Collect[{eqn3, eqn4, eqn5, eqn6, eqn7} /. values, C[_], 
   Simplify];
{knowns, mat} = CoefficientArrays[eqs, Array[C, 5]]
Det[mat] // N
eqs // N // Simplify // Chop
Solve[eqs, Array[C, 5]]