Answer
a) The Brackett and Paschen series overlap.
b) No overlap of Balmer series with Paschen series.
Work Step by Step
(a) We know that for $n_1=3$, $n=\infty$ and $n_1=4$ and $n=\infty$, we have
$\lambda_{\infty}=\frac{1}{(1.097\times 10^7)(\frac{1}{3^2})-\frac{1}{\infty^2}}=8.204\times 10^{-7}m=820nm$
Similarly $\lambda_{\infty}=\frac{1}{(1.097\times 10^7)(\frac{1}{4^2}-\frac{1}{\infty^2})}=14.585\times 10^{-7}m=1458.5nm$
Thus, $820nm\lt \lambda_{Paschen}\lt 1875nm$
$\space 1458.5nm\lt \lambda_{Brackett}\lt 4051nm$
We conclude that the Brackett and Paschen series overlap.
(b) We know that for Balmer series and Paschen series
$\lambda_3=\frac{1}{(1.097\times 10^7)(\frac{1}{2^2}-\frac{1}{3^2})}=0.6563\mu m$
and $\lambda_4=\frac{1}{(1.097\times 10^7)(\frac{1}{3^2}-\frac{1}{4^2})}=1.875\mu m$
Now, we check the above both scenarios for $n=\infty$
$\lambda_{\infty}=\frac{1}{(1.097\times 10^7)(\frac{1}{2^2}-\frac{1}{\infty^2})}=364.6nm$
and $\lambda_{\infty}=\frac{1}{(1.097\times 10^7)(\frac{1}{3^2}-\frac{1}{\infty^2})}=0.8204\mu m$
Thus, $364.6nm\lt \lambda_{Balmer}\lt 656.33nm \space 820.4nm\lt \lambda_{Paschen}\lt 1875nm$
We conclude that there is no overlap of Balmer series with Paschen series.