Let $f:X \to Y$ be a projective morphism of complex Noetherian schemes. Assume $Y$ is smooth and for all $y \in Y$, $f^{-1}(y)$ is of pure dimension $1$. Let $\mathcal{F}_1, \mathcal{F}_2$ and $\mathcal{F}_3$ be coherent sheaves on $X$ flat over $Y$ satisfying the following short exact sequence:

$$0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0.$$

Assume $R^0f_*(\mathcal{F}_1)$ and $R^1f_*(\mathcal{F}_3)$ are zero. We then have the associated long exact sequence of right derived functor:

$$0 \to R^0f_*(\mathcal{F}_2) \to R^0f_*(\mathcal{F}_3) \to R^1f_*(\mathcal{F}_1) \to R^1f_*(\mathcal{F}_2) \to 0.$$

Assume $R^0f_*(\mathcal{F}_3)$ and $R^1f_*(\mathcal{F}_1)$ are locally free $\mathcal{O}_Y$-modules. Then,

$1)$ Is it possible that one of $R^0f_*(\mathcal{F}_2)$ or $R^1f_*(\mathcal{F}_2)$ is locally free $\mathcal{O}_Y$-module?

$2)$ For any $y \in Y$ closed, denote by $k(y)$ the constant sheaf $k(y)$ on the point $y$ (as mentioned in Hartshorne's Algebraic geometry Corollary, III.$9.4$). Is it then true that $- \otimes k(y)$ (as mentioned in loc. cit. Theorem III.$12.11$) is exact when applied to the above long exact sequence. In particular, is the following sequence exact?

$$0 \to R^0f_*(\mathcal{F}_2)\otimes k(y) \to R^0f_*(\mathcal{F}_3)\otimes k(y) \to R^1f_*(\mathcal{F}_1)\otimes k(y) \to R^1f_*(\mathcal{F}_2)\otimes k(y)$$