Consider the ideal $I=(a)\cap(a,b)^2=(a^2,ab)$ in $k[a,b]$ and set $R=k[a,b]/I$. The problem is to show that for all $n\ge 1$ and all $\lambda\ne 0$, the ideals $(b^n)$ and $(a+\lambda b^n)$ of $R$ are $(a,b)$-primary in $R$. This amounts to saying that the set of associated primes of the quotient $k[a,b]$-modules $R/(b^n)$ and $R/(a+\lambda b^n)$ is the singleton consisting of the ideal $(a,b)\subset k[a,b]$.
First, it would be reasonable to find "simpler" $k[a,b]$-modules isomorphic to $R/(b^n)$ and $R/(a+\lambda b^n)$, but I'm not sure how to do that. I don't see how one can apply any of the isomorphism theorems, for example. It is plausible to think that $R/(a+\lambda b^n)\simeq k[b]/(b^{n+1})$ since $b^{n+1}=b(b^n)=b(-a/\lambda)=-(ab)/\lambda=0$ in the quotient, but I don't see how to prove this formally.
Even if I do know that last isomorphism, I'm not sure how to find $\operatorname{Ass}$ of that module.
Added:
According to the hint in the answers, $R/(b^n)\simeq k[a,b]/(a^2,ab,b^n)$, and according to my conjecture, $R/(a+\lambda b^n)\simeq R[b]/(b^{n+1})$. Now $(a,b)$ is an associated prime of each of the quotient modules because $k[a,b]/(a,b)$ injects into them (added later: is that true at all? Originally I thought that this is true because $k[a,b]/(a,b)\simeq k$ and any ring homomorphism from a field to a ring is injective. But here we are dealing with module homomorphisms...). According to Theorem 3.10 from Eisenbud, to show that $(a,b)$ is the only associated prime of those quotient modules, it suffices to show that the $k[a,b]$-submodule $(a)$ of $R$ is also $(a,b)$-primary in $R$ and
$$(0)=(a)\cap (b^n)\\ (0)=(a)\cap (a+\lambda b^n)$$
are minimal primary decompositions.
Well, if there is a smaller primary decomposition, then $(0)$ is a primary ideal of $R$. But I don't see what it contradicts too. Also, to prove that $(a)$ is $(a,b)$ primary, we need to show (as above) that $k[a,b]/(a,b)$ injects into $R/(a)$. Here I have the same question as with the other injections above.
Hint:
Whether a,b denote indeterminates or not, one has the isomorphism: \begin{align}R/(b^n)&=k[a,b]/(a^2,ab)\bigm/b^n\!\cdot k[a,b]/(a^2,ab)=k[a,b]/(a^2,ab)\!\Bigm/\!(b^n,a^2, ab)/(a^2,ab)\\ &\simeq k[a,b]/(b^n,a^2, ab)\;\qquad\text{ (3rd isomorphism theorem)} \end{align}