Is the category of these particularly nice spaces cartesian closed? Is the category of Hausdorff, compactly generated, locally path-connected, semi-locally 1-connected spaces (and continuous maps between them) cartesian closed? If not, in what ways does it fail to be?
 A: I don't think $Y^X$ is semilocally simply connected when $X$ isn't compact. The argument below is incomplete in that it uses the standard compact open topology on $Y^X$, while one would at least have to $k$-ify it, and probably do some more to stay inside the category you are considering. Nevertheless, I think it shows that semilocal simple connectedness probably fails for function spaces.
Consider $X=Y$ equal to the disjoint union of infinitely many circles $$\coprod_{n\in\Bbb N}\;S^1$$ and set $f=\mathrm{id}\in Y^X$. Semilocal simple connectedness means that there is a neighborhood $U\subset Y^X$ of $f$ with the property that every loop drawn in $U$ (and based at $f$) can be shrunk to a point inside $Y^X$. Equivalently, that there is a small neighborhood of $f$ with the property that every homotopy $H_t:f\xrightarrow{\sim} f$ drawn inside $U$ is homotopic to the constant homotopy inside $Y^X$. A neighborhood of $f$ in the compact open topology always contains an intersection of subbasic neighborhoods, say $U'$. Being a map $X\to Y$ in $U'$ simply means that you are close to $f$ on some compact subset $K$ of $X$. By compactness, $K\subset\coprod_{0\leq n<N}\;S^1$ for some $N$.
Then consider the homotopy $H$ between $f$ and $f$ with $H_t$ being the identity on all circles except for the $N$th circle where $H_t$ is rotation by $2\pi t$. By definition of $N$, this loop $H_t:f\to f$ stays inside $U'$. If $X^Y$ were semilocally simply connected there'd be a homotopy of loops from $H$ to the constant loop at $f$, and upon restricting to the $N$th circle we'd get a homotopy inside $(S^1)^{S^1}$ from the loop $t\mapsto R_{2\pi t}$ (rotation by $2\pi t$)  to the constant loop $t\mapsto f$. Applying this homotopy of maps at a specific point, for instance $1\in S^1$ would give a homotopy from the identity of $S^1$ to the constant map $S^1\to S^1,z\mapsto 1$, which doesn't exist.
