Question 1. Does every Banach space admit a strictly convex equivalent norm?
Not in general. In fact, in [Chapter II.7, 1] it is shown that
$$ \text{for $\Gamma$ an uncountable set, $\ell_{\infty}(\Gamma)$ admits no equivalent strictly convex norm.}$$
As noted by the OP, this fails for $\Gamma=\mathbb N$ since $\ell^\infty(\mathbb N)$ does admit an equivalent strictly convex norm. Day [3] showed that
Theorem. If there exists an injective bounded linear operator
$$T \colon X \to c_0(\Gamma)$$
for some set $\Gamma$, then $X$ admits an equivalent strictly convex norm.
Here, $c_0(\Gamma)$ is the set of all bounded real functions $f$ such that for every $ε>0$ the set $(|f|>ε)$ is finite.
$\ell_\infty(\mathbb N)$ falls into that category with $\Gamma = \mathbb N,$ $c_0(\mathbb N) = \{ (x_n) \colon x_n \to 0\}$ and
$$T((x_n)_{n=1}^\infty)=(\tfrac 1n x_n)_{n=1}^\infty.$$
However, for separable spaces this is indeed the case:
Fact. Any separable Banach space $X$ has an equivalent strictly convex norm.
This does not extend to the non separable case, as shown by the example of $\ell_\infty(\Gamma)$ with $\Gamma$ uncountable.
The proof is quite simple (compare with your example in $c_0$):
Let $(x_n)$ be a dense set in $X$. Use Hahn-Banach to find $f_n \in X^*$ with norm one such that $f_n(x) =\|x_n\|$. Let $T \colon X \to \ell_2$ be given by
$$T(x) = \left( \tfrac {1}{2^n} f_n(x) \right)_{n=1}^\infty \in \ell_2, \quad x \in X.$$
Then $T$ is bounded and injective. Then $|x|:= \|x\|+\|Tx\|$ is an equivalent strictly convex norm. Hint: To see this, one can use the fact that $(X,|\cdot|)$ is strict convexity iff $|x+y|=|x|+|y|$ implies that $x=ay$ for some $a>0$.
In fact, much more can be said in the separable case:
Theorem (7.1 in [1]). Any separable Banach space admits an equivalent norm that (among other things) is Locally Uniformly Convex (LUC) i.e.,
$$ \| x_n\| = \|x\|=1 \text { and } \|x_n +x\| \to 2 ~ \text{ imply } ~ \|x_n-x\| \to 0.$$
In the (stronger) case that $X^*$ is separable then $X$ admits an equivalent norm such that both $X$ and $X^*$ are LUC.
and in the reflexive case:
Theorem (Troyanski [4]). If $X$ is reflexive, then both $X$ and $X^*$ admit an equivalent LUC (and Frechet differentiable) norm.
Question 2. Does every Banach space admit an equivalent uniformly convex (UC) norm?
Again, the answer is no. In fact, there is a characterization of such spaces proved by Enflo [2]: they are precisely the super-reflexive spaces. Super-reflexivity is a stronger notion of reflexivity that, essentially, means that every space $E$ that is "locally represented" in $X$ must be reflexive (one can take $E=X^{**}$ to infer that super-reflexivity implies reflexivity).
See this for an explicit example of a reflexive space which is LUC but not UC.
[1] Smoothness and Renormings in Banach Spaces
[2] Banach spaces which can be given an equivalent uniformly convex norm
[3] Day, M. M. (1955). Strict Convexity and Smoothness of Normed Spaces. Transactions of the American Mathematical Society, 78(2), 516–528.
[4] S . L . T ROYANSKI , ‘On locally uniformly convex and dif ferentiable norms in certain non-separable Banach spaces’, Studia Math . 37 (1971) 173 – 180 .