let me work out the comments a bit further, starting from the identity
$$\int_{{\rm O}(n)}{\rm tr}\,g^p\,dg=\begin{cases} 0&{\rm if}\;\;p\;\;\text{is an odd integer}\\ 1&{\rm if}\;\;p\;\;\text{is an even integer} \end{cases} $$ the result for $p$ odd is obvious because contributions $\pm g$ to the integral cancel; the result for $p$ even follows without calculation from the fact that only eigenvalues $\pm 1$ of the orthogonal matrix $g$ contribute to the average of the trace (the complex eigenvalues average out to zero); for $n$ odd every $g$ has one such eigenvalue, so the average is one, for $n$ even and ${\rm det}\,g=+1$ there are no such eigenvalues but for ${\rm det}\,g=-1$ there are two, so the average is again one.
so the second integral of the OP evaluates to $$\int_{{\rm O}(n)}g^p\,dg=\begin{cases} 0&{\rm if}\;\;p\;\;\text{is an odd integer}\\ \mathbb{1}&{\rm if}\;\;p\;\;\text{is an even integer} \end{cases} $$
Taylor expansion of the exponent in the first integral of the OP and term-by-term integration gives $$\int_{{\rm O}(n)}e^g\,dg=\mathbb{1}\sum_{n=0}^\infty \frac{1}{(2n)!}=\mathbb{1}\cosh 1 $$