$\newcommand{\la}{\lambda}$Up to the equality in distribution, 
\begin{equation*}
	Y=\sqrt{\sum_1^k(Z_i+L_k)^2},
\end{equation*}
where $Z,Z_1,\dots,Z_k$ are independent standard normal random variables (r.v.'s) and 
\begin{equation*}
	L_k:=\la/\sqrt k\to L
\end{equation*}
 (as $k\to\infty$).  

By the law of large numbers,
\begin{equation*}
	\frac{Y^2}k=\frac1k\sum_1^k(Z_i+L)^2+(L_k-L)^2+2(L_k-L)\frac1k\sum_1^k (Z_i+L) \\ 
	\to E(Z+L)^2=1+L^2  
\end{equation*}
and hence 
\begin{equation*}
	\frac{Y}{\sqrt k}\to\sqrt{1+L^2}   
\end{equation*}
in probability and hence in distribution. 

Also, $Var((Z+L_k)^2)=2+4L_k^2$ and hence $Var(Y^2)=k\,Var((Z+L_k)^2)=2k+4\la^2\sim2k(1+2L^2)$. 
So, by the central limit theorem (or, more precisely, by, say, the Berry--Esseen inequality), 
\begin{equation*}
	\frac{Y^2-(k+\la^2)}{\sqrt{2k(1+2L^2)}}\to Z\sim N(0,1) 
\end{equation*}
in distribution. 
So, by Slutsky's theorem,
\begin{equation*}
	Y-\sqrt{k+\la^2}=R_k:=\frac{(Y^2-(k+\la^2))/\sqrt{2k}}{(Y+\sqrt{k+\la^2})/\sqrt{2k}}
	\to\frac{\sqrt{1+2L^2}}{\sqrt{2(1+L^2)}} \,Z
\end{equation*}
in distribution. 

Also, by (say) the Rosenthal inequality, 
\begin{equation*}
	ER_k^4\le E\Big(\frac{(Y^2-(k+\la^2))/\sqrt{2k}}{(\sqrt{k+\la^2})/\sqrt{2k}}\Big)^4=O(1). 
\end{equation*}
So, by uniform integrability, 
\begin{equation*}
	E(Y-\sqrt{k+\la^2})\to E\frac{\sqrt{1+2L^2}}{\sqrt{2(1+L^2)}} \,Z=0 
\end{equation*}
and 
\begin{equation*}
	E(Y-\sqrt{k+\la^2})^2\to E\Big(\frac{\sqrt{1+2L^2}}{\sqrt{2(1+L^2)}} \,Z\Big)^2=\frac{{1+2L^2}}{2(1+L^2)}.  
\end{equation*}
Thus, 
\begin{equation*}
	EY=\sqrt{k+\la^2}+o(1) \tag{1}\label{1}
\end{equation*}
and 
\begin{equation*}
Var\, Y=	E(Y-\sqrt{k+\la^2})^2-(E(Y-\sqrt{k+\la^2}))^2\to\frac{{1+2L^2}}{2(1+L^2)}.  \tag{2}\label{2}
\end{equation*}