Ans (a)$467856$ (b)$7485696$

For (a) if $\gcd(x,y)=1$ it means $x^2y^2\mid k$. So let $k=x^2y^2m$, using the hypothesis we get $$y^2m=n,\qquad x^2m=n+148$$In particular we have $$\frac{n}{y^2}=\frac{n+148}{x^2}\iff \frac{x^2}{y^2}=\frac{n+148}{n}$$Now let $\gcd(148,n)=t$ so $148=ta$ and $n=tb$ for $\gcd(a,b)=1$. Thus $$\frac{x^2}{y^2}=\frac{a+b}{b}$$i.e since $\gcd(x,y)=1$ and $\gcd(a+b,b)=1$ we have $x^2=a+b$ and $y^2=b$. Notice we have $(x-y)(x+y)=a$ and $a\mid 148$ So we get $a=37$ and $x=19,~y=18$ or $a=148$ and $x=38,~y=36$ moreover the latter contradicts form of $x,y$ so we are done. We get $b=324$ and $t=4$ so $m=\frac{4\cdot 324}{18^2}=4$ thus $k=19^2\cdot 18^2\cdot 4=467856$