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
$$\int_{{\rm O}(n)}g^p\,dg=\begin{cases}
0&{\rm if}\;\;p\;\;\text{is an odd integer}\\
n^{-1}\mathbb{1}&{\rm if}\;\;p\;\;\text{is an even integer}
\end{cases}
$$

Taylor expansion of the exponent in the first integral of the OP and term-by-term integration gives
$$\int_{{\rm O}(n)}e^g\,dg=n^{-1}\mathbb{1}\sum_{p=0}^\infty \frac{1}{(2p)!}=\frac{\cosh 1}{n}\mathbb{1}
$$

now for the third integral of the OP, we need the fourth-order tensor
$$\int_{{\rm O}(n)}(g^p)_{ij}(g^q)_{kl}\,dg=a_{pq}(n)\delta_{ij}\delta_{kl}+b_{pq}(n)\delta_{ik}\delta_{jl}+c_{pq}(n)\delta_{il}\delta_{jk}$$
so that the required integral takes the form
$$\int_{{\rm O}(n)}g^p Ag^q\,dg=a_{pq}(n)A+b_{pq}(n)A^{\rm t}+c_{pq}(n)\mathbb{1}\,{\rm tr}\,A
$$
by taking the trace I know that
$$a_{pq}(n)+b_{pq}(n)+nc_{pq}(n)=\begin{cases}
0&{\rm if}\;\;p+q\;\;\text{is an odd integer}\\
1&{\rm if}\;\;p+q\;\;\text{is an even integer}
\end{cases}
$$
two more relations are needed...