Let $M$ be a closed (compact without boundary) Riemannian manifold. Let $p,q\in M$ and $a\in M$ such that $d(a,p)=d(a,q)$. Does there always exists a (smooth) geodesic joining $p$ and $q$ passing through $a$?
This does not hold if the manifold is non-compact. For example, if we take $M=\mathbb{R}^2$ and $p=(-1,0)$ and $q=(1,0)$, then $a=(0,1)$ does not satisfy the above property.
No.
Take the unit sphere in $\mathbb R^3$.
Let $p = (-\epsilon, \sqrt{1-\epsilon^2}, 0)$ and $q=(\epsilon, \sqrt{1-\epsilon^2}, 0)$ where $\epsilon$ is small. Then there is a unique geodesic that passes through $p$ and $q$, i.e. the circle $x^2+y^2=1, z=0$.
Meanwhile, the points $a$ with $d(a,p)=d(a,q)$ form the circle $x=0, y^2+z^2=1$.