A. I wonder if every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ is a square element in ${\rm Spin}_n^{\epsilon}(q)$?

B. When $Z(\Omega_{2m}^{\epsilon}(q))\cong C_2$, is the unique element of order two in $Z(\Omega_{2m}^{\epsilon}(q))$ a square element in $\Omega_{2m}^{\epsilon}(q)$？

Notes： 

1. Here the ground field is a finite field $F_q$ with $q$ a power of some prime $p$.

2. We always set $n=2m\geq 6$ and $q^m\equiv \epsilon~({\rm mod}~4)$.

Some quoted results:

If $m\geq 3$, then $P\Omega^{\epsilon}(2m,q)$ is a finite simple group.

It is easy to find elements of the spin group which square to $-1$, and hence the spin group is a proper double cover of the orthogonal group. We write ${\rm Spin}_n^\epsilon(q)$ for this group of shape $2.\Omega_n^\epsilon(q)$.

If $n$ is odd, or if $n=2m$ and $q^m\equiv -\epsilon~{\rm mod}~4$, then $\Omega_n^\epsilon(q)$ is already simple and the spin group has the structure $2.\Omega_n^\epsilon(q)$.

If $n=2m$ and $q^m\equiv \epsilon~({\rm mod}~4)$, then $\Omega_n^\epsilon(q)$ has a centre of order 2, and the spin group has the structure $4.{\rm P\Omega}_n^\epsilon (q)$ if $m$ is odd, and the structure $2^2.{\rm P\Omega}_n^\epsilon (q)$ (necessarily with $\epsilon=+$) if $m$ is even.

$P\Omega_{2m}^+(q)=D_m(q)$ for $m\geq 3$ and its Schur multiplier is $C_{(4, q^m-1)}$ if $m$ is odd and 

$$C_{(2, q^m-1)}\times C_{(2,q^m-1)}$$ 

if $m$ is even.

$P\Omega_{2m}^{-}(q)={}^2D_m(q)$ for $m\geq 2$ and its Schur mulitiplier is $C_{(4, q^m+1)}$.

$P\Omega_6^+(q)=PSL_4(q)$. 

$P\Omega_6^-(q)=PSU_4(q)$.

Let $P$ be a $2$-group of cyclic center $\langle a \rangle$ and let $\omega_n(P)=P\wr C_2\wr C_2\cdot\cdot\cdot \wr C_2$ be the wreath product of $P$ and $n$ copies of $C_2$, where $n\geq 2$.

Suppose that $2^{t+1}$ is the $2$-part of $q^2-1$. Let $T$ be a central product of two dihedral groups 

$$D_1=\langle d, h: h^{-1}dh=d^{-1}\rangle$$ 

and 

$$D_2=\langle g, k: k^{-1}gk=g^{-1}\rangle$$ 

of order $2^{t+1}(d^{2^{t-1}}=g^{2^{t-1}})$ and let $e, f\in {\rm Aut}T$ be chosen such that 

$$o(e)=o(f)=2, [e,f]=1,$$ 

$$d^e=g^{-1}, g^e=d^{-1}, h^e=gk, k^e=dh, d^f=g, g^f=d, h^f=k, k^f=h.$$ 

The twisted wreath product $tw_1(C)$ of $T$ and $C$, where

$$C=\langle 
\left(
  \begin{array}{cc}
    e_2 & 0 \\
    0 & e_2 \\
  \end{array}
\right), 
\left(
  \begin{array}{cc}
    f_2 & 0 \\
    0 & f_2 \\
  \end{array}
\right), 
\left(
  \begin{array}{cc}
    0 & I_2 \\
    I_2 & 0 \\
  \end{array}
\right)
\rangle$$

($e_2$ and $f_2$ are $2\times 2$ diagonal matrices of the forms ${\rm diag}(e,1)$ and ${\rm diag}(f,1)$), is the group

$$tw_1(T)=\langle \pmatrix{T & 0\\ 0 & I_2}, \pmatrix{I_2 & 0\\ 0 & T}, \pmatrix{e_2 & 0\\ 0 & e_2}, \pmatrix{f_2 & 0\\ 0 & f_2}, \pmatrix{0 & I_2\\ I_2 & 0}\rangle.$$

Note that $C$ is elementary Abelian of order 8. In general, $tw_{n+1}$ (the twisted wreath product of $T$ and $n+1$ copies of $C$) is generated by

$$\langle U=\pmatrix{tw_n(T) & 0\\ 0 & I_{2^n}}, V=\pmatrix{I_{2^n} & 0\\ 0 & tw_n(T)}\rangle\cong tw_n(T)\times tw_n(T)$$

and

$$\langle \pmatrix{e_{2^n} & 0\\ 0 & e_{2^n}}, \pmatrix{f^{2^n} & 0\\ 0 & f^{2^n}}, \pmatrix{0 & I_{2^n}\\ I_{2^n} & 0}\rangle\cong C$$

where $e_{2^n}$ and $f_{2^n}$ are $2^n\times 2^n$ diagonal matrices of the form

$${\rm diag}(e, 1, ..., 1)$$ 

and 

$${\rm diag}(f, 1, ..., 1).$$ 

Let $z$ be the generator of the center of $T$ and let 
$$E=\prod_{a\in tw_{n+1}(T)}\langle z^a\rangle.$$
Then $E$ is elementary Abelian of order $2^{n+1}$. Suppose that $E=\prod\langle z_i\rangle$ (direct product). Then $z_0=\prod z_i$ generates the center of $tw_{n+1}(T)$. 

$\omega_{n-2}(T)$ is a Sylow $2$-subgroup of $\Omega^{\epsilon}(2^n,q)$. Further, $\omega_{n-2}(T)/Z$, where $Z=\langle z_0\rangle$, is a Sylow 2-subgroup of $P\Omega^\epsilon(2^n,q)$.

Let $S$ be a Sylow $2$-subgroup of $P\Omega^\epsilon(2m,q)$, where $m\geq 4$, $q$ is a power of an odd prime and $q^m\equiv \epsilon (\rm mod~4)$. Let $F_q$ be the field of $q$ elements. Let $\Phi_1$ be the determinant mapping and $\Phi_2$ be the spinorial norm mapping $\Phi: O^\epsilon\rightarrow F_q^\times/{F_q^\times}^2\cong C_2$. It is clear that 

$${\rm ker \Phi_1}\cap {\rm ker \Phi_2}=P\Omega^\epsilon(2m,q)$$ 

Denote by 
$Z$ the center of $O^\epsilon(2m,q)$. Define $\phi_i=\Phi_i|_T$. Let $S'={\ker \phi_1}\cap {\rm ker \phi_2}$. Then $S'$ is a Sylow $2$-subgroups of $\Omega^{\epsilon}(2m,q)$. Since th determinant and spinorial norm of members in $Z$ are $1$ and perfect squares respectively, $Z\leq S'={\rm ker \phi_1}\cap {\rm ker \phi_2}$.

$Lie(r)$ is the set of finite groups possessing a $\sigma$-setup $(\bar{K},\sigma)$ over $\bar{F}_r$ such that $\bar{K}$ is simple. Furthermore,
$$Lie=\bigcup_r Lie(r),~~~{\rm the~union~over~all~primes}~r$$

If $\sum=D_{2m}$, then $Z(\bar{K}_u)$ is $\bar{F}^{(2)}\times \bar{F}^{(2)}$.

If $\sum=D_{2m}$, then the generators of $Z({\bar{K}})$ are $h_1=h_{\alpha_1}(-1)h_{\alpha_3}(-1)...h_{\alpha_{2m-1}}(-1)$ and $h_2=h_{\alpha_{2m-1}}(-1)h_{\alpha_{2m}}(-1)$.

Analysis:

Let

$$\pi: \Omega_6^-(3)\rightarrow P\Omega_6^-(3).$$

If $x^2=-1$, then $f(x)$ is an element of order 2, however $o(x)=4$, a contradiction by Richard Lyons's notes below.

The following websites may be useful to my question.

https://mathoverflow.net/questions/140568/double-covers-of-the-orthogonal-groups#

http://brauer.maths.qmul.ac.uk/Atlas/v3/