If $Y\subset X$ are Banach spaces such that $Y$ is dense in $X$, is it true that $X'$ is dense in $Y'$?

If $$Y$$ is a dense subspace of a Banach space $$(X,\|\cdot\|_1)$$ and $$(Y,\|\cdot\|_2)$$ is a Banach space such that the inclusion from $$(Y,\|\cdot\|_2)$$ into $$(X,\|\cdot\|_1)$$ is continuous, then it is well defined, linear, injective, and continuous in the dual norm topology the map: $$j:X'\to Y', f\mapsto f|_{Y},$$ where $$X'$$ is the topological dual of $$(X,\|\cdot\|_1)$$ and $$Y'$$ is the topological dual of $$(Y,\|\cdot\|_2)$$.

So, we can identify $$X'$$ as a subset of $$Y'$$.

Is it true that $$X'$$ is dense in $$Y'$$ in the norm topology? If not, is true that $$X'$$ is dense in $$Y'$$ at least in the weak* topology?

Edit: in this question it is addressed the case where $$(Y,\|\cdot\|_2)$$ is reflexive, obtaining that in this case (thanks to Hahn-Banach theorem) $$X'$$ is dense in the norm topology of $$Y'$$. In the answer to this question it is shown that counterexamples to density in norm topology exist if the reflexivity of $$(Y,\|\cdot\|_2)$$ is not assumed, e.g. by taking $$(X,\|\cdot\|_1):=(l^2,\|\cdot\|_{l^2})$$ and $$(Y,\|\cdot\|_1):=(l^1,\|\cdot\|_{l^1})$$. However, in this counterexample $$X'$$ is still dense in the weak* topology of $$Y'$$.

So, it remains to answer only the following part of the original question:

Is true that $$X'$$ is dense in $$Y'$$ in the weak* topology?

With this $G$ is dense in $X^*$ in weak* sense if and only if $G$ is total set fact, the proof is rather easy.
Let me denote the embedding of $$Y$$ into $$X$$ by $$i$$. Then $$i(Y)$$ is dense in $$X$$. And the question is, whether $$i^*(X')$$ is weak-star dense in $$Y'$$.
Due to the result above, we have this density if and only if for all $$y\in Y$$ $$(i^*f)(y) =0 \quad \forall f\in X' \Rightarrow y=0.$$ Now let $$y\in Y$$ be given such that $$(i^*f)(y)=f(iy)=0$$ for all $$f\in X$$. This implies $$iy=0$$, and by injectivity of $$i$$, $$y=0$$. So $$i^*X'$$ is total and hence dense in $$Y'$$.
Example. Sending each sequence to itself, we get $$l^1 \longrightarrow c_0$$ injective with dense range. Taking adjoint, we get $$l^\infty \longleftarrow l^1$$ injective but non-dense range.