For completeness, here is the same solution using NeumannValue:

s1 = NDSolve[{D[x[n, t], t] == 
    D[x[n, t], {n, 2}] - q[n, t] x[n, t] + 
     NeumannValue[0.005, n == 0] + NeumannValue[0.005, n == 90], 
   D[q[n, t], t] == -q[n, t] Exp[x[n, t]], 
   x[n, 0] == Sin[(n + 90/4)* 2*Pi/90] + 0.005*n, 
   q[n, 0] == 0.01}, {x[n, t], q[n, t]}, {n, 0, 90}, {t, 0, 10}]
Plot3D[x[n, t] /. s1, {n, 0, 90}, {t, 0, 10}, PlotRange -> All]
Plot3D[(x[n, t] /. s) - (x[n, t] /. s1), {n, 0, 90}, {t, 0, 10}, 
 PlotRange -> All]

The last plot compares (subtracts) the two solutions to get zero.

Shruti,

You need to do this:

Plot[(x[n, t] /. s /. n -> 20) , {t, 0, 10}, PlotRange -> All]

You want to plot x[n,t] using the solution, s:

x[n,t] /. s

s has a substitution rule for x and q in the form of an interpolation function. you need to apply that rule to tell Mathematica what solution you want, x[n,t] or q[n,t]. After that you can substitute in values for n and/or t using /. n->20 or /. t-> 5, etc.

Regards,

Neil