Answer: (a) For \(n \geq 2\), \(I_n=-1+n(n-1)I_{n-2}\).

(b) \(I_2=2\cosh 1-3\), equivalently \(I_2=\mathrm e+\mathrm e^{-1}-3\).

(a) Start with

\(I_n=\int_0^1 (1-x)^n\sinh x\,\mathrm dx\).

Integrate by parts with \(u=(1-x)^n\) and \(\mathrm dv=\sinh x\,\mathrm dx\). Then \(\mathrm du=-n(1-x)^{n-1}\,\mathrm dx\) and \(v=\cosh x\), so

\(I_n=\left[(1-x)^n\cosh x\right]_0^1+n\int_0^1(1-x)^{n-1}\cosh x\,\mathrm dx\).

Now

\(\left[(1-x)^n\cosh x\right]_0^1=0^n\cosh 1-1^n\cosh 0= -1\),

so

\(I_n=-1+n\int_0^1(1-x)^{n-1}\cosh x\,\mathrm dx\).

Integrate by parts again on the remaining integral, with \(u=(1-x)^{n-1}\) and \(\mathrm dv=\cosh x\,\mathrm dx\). Then \(\mathrm du=-(n-1)(1-x)^{n-2}\,\mathrm dx\) and \(v=\sinh x\), giving

\(\int_0^1(1-x)^{n-1}\cosh x\,\mathrm dx=\left[(1-x)^{n-1}\sinh x\right]_0^1+(n-1)\int_0^1(1-x)^{n-2}\sinh x\,\mathrm dx\).

Since \(n\ge 2\), the boundary term is

\(\left[(1-x)^{n-1}\sinh x\right]_0^1=0^{n-1}\sinh 1-1^{n-1}\sinh 0=0\).

Hence

\(\int_0^1(1-x)^{n-1}\cosh x\,\mathrm dx=(n-1)I_{n-2}\).

Substituting back,

\(I_n=-1+n(n-1)I_{n-2}\).

(b) First find \(I_0\):

\(I_0=\int_0^1\sinh x\,\mathrm dx=\left[\cosh x\right]_0^1=\cosh 1-1\).

Using the recurrence with \(n=2\),

\(I_2=-1+2I_0=-1+2(\cosh 1-1)=2\cosh 1-3\).

So the exact value is \(\boxed{2\cosh 1-3}\).