Here is a way to answer the second question. I am assuming complex
matrices and using '$X^{*}=\overline{X}^{\mathrm{T}}$. The first question (already answered by Suvrit ) I discuss at the end.

Assume '$U^{*}U=I$ and $U^{\mathrm{T}}=-U$. If $\mathbf{v}$ is nonzero and $U\mathbf{v}=\lambda\mathbf{v}$ then $|\lambda|=1$ and the spectral theorem for normal matrices implies $U^{*}\mathbf{v}=\overline{\lambda}\mathbf{v}$. Take the
conjugate to find $U^{\mathrm{T}}\overline{\mathbf{v}}=\lambda\overline{\mathbf{v}}$. Since $U^{\mathrm{T}}=-U$ we find $U\overline{\mathbf{v}}=-\lambda\overline{\mathbf{v}}$. This mean $\overline{\mathbf{v}}$ and $\mathbf{v}$ are orthogonal unit vectors. We could have started with an orthonormal basis for the whole eigenspace of $\lambda$ and so can find an orthonormal basis $\mathbf{v}_{1},\overline{\mathbf{v}_{1}},\dots,\mathbf{v}_{N},\overline{\mathbf{v}_{N}}$ so we see we are on an even dimensional space. Let the corresponding eigenvalues be $\lambda_{1},-\lambda_{1},\dots,\lambda_{N},-\lambda_{N}$.

We get a new real orthonormal basis $\mathbf{w}_{1},\dots,\mathbf{w}_{2N}$
by defining
$$
\mathbf{w}_{2j}=\frac{1}{\sqrt{2}}\mathbf{v}_{j}+\frac{1}{\sqrt{2}}\overline{\mathbf{v}_{j}}
$$
and
$$
\mathbf{w}_{2j+1}=\frac{i}{\sqrt{2}}\mathbf{v}_{j}-\frac{i}{\sqrt{2}}\overline{\mathbf{v}_{j}}
$$
Since
$$
U\mathbf{w}_{2j}=\lambda_{j}\frac{1}{\sqrt{2}}\mathbf{v}_{j}-\frac{1}{\sqrt{2}}\lambda_{j}\overline{\mathbf{v}_{j}}=-i\lambda_{j}\mathbf{w}_{2j+1}
$$
and
$$
U\mathbf{w}_{2j+1}=\lambda_{j}\frac{i}{\sqrt{2}}\mathbf{v}_{j}+\lambda_{j}\frac{i}{\sqrt{2}}\overline{\mathbf{v}_{j}}=i\lambda_{j}\mathbf{w}_{2j}
$$
we see there is a real orthogonal matrix $O$ so that
$$
U=O^{*}DO=O^{\mathrm{T}}DO
$$
where $D$ is block diagonal with blocks
$$
\left[\begin{array}{cc}
0 & i\lambda_{j}\\
-i\lambda_{j} & 0
\end{array}\right].
$$
Let $Q$ be block diagonal with blocks
$$
\sqrt{i\lambda_{j}}\left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right]
$$
and $S$ be block diagonal with blocks
$$
\left[\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right].
$$
Then $D=QSQ=Q^{\mathrm{T}}SQ$ and
$$
U=O^{\mathrm{T}}Q^{\mathrm{T}}SQO.
$$
The desired unitary is $QO$.

If $S$ is not what you consider the standard symplectic matrix, a permutation matrix will fix that.

In the case of $U$ unitary and $U^\mathrm{T} = U$ the proof starts the same, but we get $U\overline{\mathbf{v}}=\lambda\overline{\mathbf{v}}$ This leads to real eigenvectors, and so a real orthogonal matrix $O$ with $U=O^{\mathrm{T}}DO$ and $D$ diagonal. So we have reduced to the case of a diagonal unitary, which is easy. Any diagonal square root will do.