The answer to the first question is YES (~~and probably the second as well ~~).

First of all, we may replace the original representation $\pi(A)\subset B(H)$ with a universal representation $\pi\oplus\sigma$, by replacing $x \in \pi(A)''$ with $x\oplus \frac{1}{k+1}\sigma(a) \in A^{**}$.
Then, Kaplansky's density theorem is upgraded to the following (which is
an easy consequence of the Hahn--Banach separation theorem).

Lemma 1: Let $z \in A^{**}$, $w \in A$, and a net $(z_i)_i$ in $A$.
If $\| z - w \| \le 1$ and $z_i \to z$ weak*, then
$$\lim_j\mathrm{dist}(w,\mathrm{conv}\{ z_i : i \geq j\})\le1$$

The advantage of using convex combination is that it can be iterated
without destroying the previously obtained approximation estimate.
From Lemma 1, one immediately obtains

Lemma 2: Let $x\in A^{**}$ and $a\in A$ be such that
$\|x\|\le1$ and $\| x - a \|\le k$. Then for any $\epsilon_1>0$,
the element $x$ is weak*-approximated by $y_1\in A$ such that
$\|y_1\|\le 1+\epsilon_1$ and $\| y_1 - a \|\le k+\epsilon_1$.

We are done once we show the approximant $y_1$ in Lemma 2 is norm-close
to an element that satisfies the exact norm inequalities:

Lemma 3: Let $y_1 \in A$ and $a \in A$ be such that $\|y_1\|\le1+\epsilon_1$, $\|y_1-a\|\le k+\epsilon_1$, and $\|a\|\le k+1$.
Then, there is $y\in A$ such that $\|y\|\le1$, $\|y-a\|\le k$, and $\|y-y_1\|\approx_{\epsilon_1}0$.
Here $\approx_{\epsilon_1}$ means that the difference is at most
$h(\epsilon_1)$ for some explicit continuous function $h\geq0$
such that $h(0)=0$.

Now Lemma 3 is proved by iterating the following approximate version and finding a suitable convergence sequence $(y_n)_n$:

Lemma 4: Let $y_1 \in A$ and $a \in A$ be such that $\|y_1\|\le1+\epsilon_1$, $\|y_1-a\|\le k+\epsilon_1$, and $\|a\|\le k+1$.
Then, for any $\epsilon_2>0$, there is $y_2\in A$ such that $\|y_2\|\le1+\epsilon_2$, $\|y_2-a\|\le k+\epsilon_2$, and $\|y_2-y_1\|\approx_{\epsilon_1}0$.

Proof of Lemma 4: By Lemma 1, it suffices to find $y_2$ in $A^{**}$
(as opposed to in $A$).
Put $\alpha=\beta=(2\epsilon_1)^{1/2}\approx_{\epsilon_1}0$.
Let $a=v|a|$ be the polar decomposition,

$p:=1_{[k+1-\alpha,k+1]}(|a|)$, and $q:=vpv^*$.
Since $ap\approx_{\epsilon_1}(k+1)vp$,
$\|y_1p\|\approx_{\epsilon_1}1$, $\| y_1p - ap \|\approx_{\epsilon_1}k$,
and $vp$ is a partial isometry,
one has $y_1p \approx_{\epsilon_1}vp$ and $y_1p \approx_{\epsilon_1} qy_1$.
Thus for $a':=ap^\perp=q^\perp a p^\perp$ (which has $\|a'\|\le k+1-\alpha$) and
$$y_2:= qvp + q^\perp((1-\beta)\frac{y_1}{\|y_1\|}+\beta\frac{a'}{\|a'\|})p^\perp$$
one has
$\| y_2 \|\le 1$ and $y_2\approx_{\epsilon_1}y_1$.
Moreover, since
\begin{align*}
\|q^\perp(y_2-a)p^\perp\|&\le\|y_1-\frac{y_1}{\|y_1\|}\|+\|(1-\beta)q^\perp y_1p^\perp+\frac{\beta}{\|a'\|}a' - a'\|\\
&\le \epsilon_1+(1-\beta)(k+\epsilon_1)+\beta(\|a'\|-1)\\
&\le k+2\epsilon_1-\alpha\beta = k,
\end{align*}
one has $\|y_2-a\|\le k$ (assuming $k>\alpha$).