The Calkin representation for Banach spaces

The answer to question 1 is yes.

Suppose that $T^*B_{X^*}$ is not compact. Since it is norm closed, there is $\epsilon >0$ and an infinite subset $S$ of $B_{X^*}$ so that $\|T^{*}x_1^*-T^{*}x_2^*\| > \epsilon$ for all $x_1^*\not= x_2^*$ in $S$. Let $x^*$ be any weak$^*$ limit point of $S$. For $\alpha$ in $\Lambda_0$ pick $x_\alpha^*$ in $S$ with $T^*x^*_\alpha \not= T^*x^*$ so that $\|x^* - x_\alpha^*\|_\alpha < 1/\dim \alpha$, where $\|z^*\|:= \|z^*_{|\alpha}\|$. Then by the choice of $\Lambda_0$, $x^*_\alpha \to x^*$ weak$^*$ and hence $\widehat{T}(x^{*}_\alpha -x^{*})_\alpha =0$, which means $\|T^*x_\alpha^* - T^*x^*\| \to 0$. Since $S$ is $\epsilon$-separated, this forces $T^*x_\alpha^* = T^*x^*$ eventually, which is a contradiction.

The answer to question 2 is yes.

The proof you already know since you proved that $T$ compact implies $\hat{T}$ is zero. (For someone who has not thought about this, it is immediate from the elementary fact that a bounded net in $X^*$ that converges to zero weak$^*$ must converge uniformly to zero on compact subsets of $X$.) So if $(x^*_\alpha)_\alpha$ is in $\widehat{X}$ with $\sup \|x_\alpha^\alpha \| \le 1$ and $x^*_\alpha \to 0$ weak$^*$, then $x^*_\alpha \to 0$ uniformly on $K$. Since the unit ball of $Y$ is contained in $K+\epsilon B_{X}$, it follows that $\|\hat{j}\| \le \epsilon$, so $C$ can be one.