Solution 1:

No. The $1$-norm and $\infty$-norm are not strictly convex. Their unit balls are polyhedra, and if you pick $x$ and $y$ to be distinct points on the same face, their convex combinations will be on the same face, too.

For example, let $x=(1,1)$ and $y=(1, -1)$. Then $\|\lambda x+(1-\lambda) y\|_\infty=1$ for all $\lambda\in[0,1]$. Another: let $x=(1,0)$ and $y=(0,1).$ Then $\|\lambda x+(1-\lambda)y\|_1 =1 $ for all $\lambda\in[0,1]$.