$\newcommand{\Ga}{\Gamma}$Your first inequality is true, for each $n:=d\ge2$.
Note that $a$ and $b$ equal, respectively, $X_{n+1}$ and $X_n$ in distribution, where
\begin{equation*}
X_n:=G/|G|,
\end{equation*}
$G=(G_1,\dots,G_n)$ is a standard Gaussian random vector in $\mathbb R^n$, and $|G|$ is the Euclidean norm of $G$. So,
\begin{equation*}
E\|X_n\|_1=n\,EY,
\end{equation*}
where $Y:=|G_1|/|G|$, so that $Y^2$ has the beta distribution with parameters $1/2,(n-1)/2$, and hence
\begin{equation*}
\frac{E\|X_n\|_1}{\sqrt n}=f_n:=\frac{\sqrt n}{\sqrt\pi}\,\frac{\Ga(n/2)}{\Ga((n+1)/2)},
\end{equation*}
and your first inequality means that
\begin{equation*}
r_n:=f_n/f_{n+1}\overset{\text{(?)}}\ge1. \tag{$*$}
\end{equation*}
Note that
\begin{equation*}
\rho_n:=\frac{r_{n+2}}{r_n}=\frac{(n+2)^{3/2}}{(n+1)^{3/2}}\,\sqrt{\frac{n}{n+3}}<1
\end{equation*}
for $n>0$, because
\begin{equation}
\rho_n^2-1=-\frac{2 n+3}{(n+1)^3 (n+3)}<0.
\end{equation}
So, $r_{n+2j}$ is decreasing in $j\in\{0,1,\dots\}$, for each $n>0$.
Also, it is easy to see that $r_n\to1$ (as $n\to\infty$). So, $r_n>1$ for all $n>0$, that is, ($*$) holds, as desired.
Concerning the second inequality, Pierre PC showed in a comment that $E\|X_n\|_1^2=1+(n-1)2/\pi$. Hence, the ratio
\begin{equation}
\frac{E\|X_n\|_1^2}{n}=\frac2\pi+\frac{1-2/\pi}n
\end{equation}
is decreasing in $n$, which means the the second inequality holds as well.
Asking multiple questions in one post is not encouraged on MathOverflow, I think.
