Let me focus on (1). It is answered in Lemma A.3 of this paper by Krannich. However, let me explain why his answers are 2-torsion. This argument also works for (2) in either interpretation.

That $H^i(Sp_{2n}(\mathbb{Z});\mathbb{Z}^{2n})$ is 2-torsion is equivalent to it being zero after inverting 2 in the coefficients. This can be proven using the "centre kills" trick. The group $Sp_{2n}(\mathbb{Z})$ has $C_2 = \{\pm id\}$ in its center, so we can take the quotient $Sp_{2n}(\mathbb{Z})/C_2$ and attempt to compute the desired cohomology groups using the Serre spectral sequence with coefficients in $\mathbb{Z}[1/2]^{2n}$ for the fibration sequence

$$BC_2 \to BSp_{2n}(\mathbb{Z}) \to B(Sp_{2n}(\mathbb{Z})/C_2),$$

But $H^*(BC_2;\mathbb{Z}[1/2]^{2n})$ vanishes; by a transfer argument it vanishes in positive degrees and is given by the invariants in degree 0. Thus the $E^2$-page of the spectral sequence vanishes and hence so does its abutment $H^*(Sp_{2n}(\mathbb{Z});\mathbb{Z}[1/2]^{2n})$.

If you care about the 2-torsion in higher degrees, it becomes more difficult. I suggest combining homological stability with polynomial coefficients with a variation of work of Djament and Vespa expressing the stable homology with twisted coefficients in terms of functor homology (e.g. this paper).