\(\frac{1}{1+a^{(n-m)}}+\frac{1}{1+a^{(m-n)}} = \frac{1}{\left(1+\frac{a^{n}}{a^{m}}\right)} + \frac{1}{\left(1+\frac{a^{m}}{a^{n}}\right)}\)

=  \(\frac{a^{m}}{\left(a^{m}+a^{n}\right)}+\frac{a^{n}}{\left(a^{m}+a^{n}\right)}\)

= \(\frac{{\left(a^{m}+a^{n}\right)}}{{\left(a^{m}+a^{n}\right)}}\)

= 1.

We know that 11² = 121.

Putting m = 11 and n = 2, we get:

(m - 1)^{n + 1} = (11 - 1)^{(2 + 1)} = 10³ = 1000.

Given Exp. = \(x^{(b-c)(b+c-a)}.x^{(c-a)(c+a-b)}.x^{(a-b)(a+b-c)}\) 

= \( x^{(b-c)(b+c)-a(b-c)}. x^{(c-a)(c+a)-b(c-a)}. x^{(a-b)(a+b)-c(a-b)}\)

= \(x^{(b^{2}-c^{2}+c^{2}-a^{2}+a^{2}-b^{2})} . x^{-a(b-c)-b(c-a)-c(a-b)}\)

= \(\left(x^{0}\times x^{0}\right)\)

= \(\left(1\times1\right) = 1\)

\(\left(x-\frac{1}{x}\right)^{2} = x+\frac{1}{x}-2\)

= (3 + 22) + \(\frac{1}{(3+22)}-2\)

= (3 + 22) + \(\frac{1}{(3+22)}-2\times \frac{{(3-22)}}{{(3-22)}}-2\)

= (3 + 22) + (3 - 22) - 2

   = 4.

Therefore \(\left(x-\frac{1}{x}\right) = 2.\)