For (1), take $X = c_0$ so that $X^* = \ell^1$, which is separable. Take $Y = \{y \in \ell^1 : \sum_i y(i) = 0\}$ which is a proper closed subspace. Since $Y$ contains all the elements of the form $e_i - e_{i+1}$, where $e_i$ are the usual basis functions, you can see that any $x \in Y^\perp$ satisfies $x(1) = x(2) = x(3) = \dots$. The only such element of $c_0$ is $0$.

For (1), take $X = c_0$ so that $X^* = \ell^1$, which is separable. Take $Y = \{y \in \ell^1 : \sum_i y(i) = 0\}$ which is a proper closed subspace. Since $Y$ contains all the elements of the form $e_i - e_{i+1}$, where $e_i$ are the usual basis functions, you can see that any $x \in Y_\perp$ satisfies $x(1) = x(2) = x(3) = \dots$. The only such element of $c_0$ is $0$.

The intuitive idea is that by Hahn-Banach, $Y$ will certainly have a nontrivial annihilator living in $X^{**}$. In this case it is the constants in $X^{**} = \ell^\infty$ which are not in $X$.

More generally, whenever $X$ is non-reflexive, you should be able to choose any $f \in X^{**} \setminus X$ and take $Y$ to be its kernel. Then if $x \in Y_\perp$, it is also in the annihilator $Y^\perp$ when considered as an element $f_x$ of $X^{**}$. The kernel of $f_x$ contains the kernel of $f$, so $f_x$ is a scalar multiple of $f$. Since $f_x \in X$, this is only possible if $f_x = 0$, i.e. $x=0$.