Here's a simple counterexample to Question 1:  Let $d=4$ and $k=2$.  Let $X\subset\mathrm{GL}_4(\mathbb{R})$ consist of a single element $J$ where $J^2=-I$.  Then $J$ is not conjugate to any upper triangular matrix (over $\mathbb{R}$) since it does not have real eigenvalues.  Meanwhile, since 
$$
\bigl(\Lambda^2(J)\bigr)^2 = \Lambda^2(J^2) = \Lambda^2(-I) = I,
$$ 
the eigenvalues of
$\Lambda^2(J):\Lambda^2(\mathbb{R}^4)\to\Lambda^2(\mathbb{R}^4)$ are $\pm 1$, and $\Lambda^2(\mathbb{R}^4)$ has a basis of $\Lambda^2(J)$-eigenvectors, so $\Lambda^2(J)$ is not only 'triangularizable', it's diagonalizable.

On the other hand, if $X$ is a *connected* Lie subgroup of $\mathrm{GL}_d(\mathbb{R})$ such that 
$$
\Lambda^k(X) = \{ \Lambda^k(x)\ |\ x\in X\ \}
$$
is conjugate to an upper triangular subgroup of $\mathrm{GL}_m(\mathbb{R})$
where $m = {d\choose k}$ and $1< k < d$, then $X$ must, itself, be conjugate to an upper triangular subgroup of $\mathrm{GL}_d(\mathbb{R})$.  This essentially follows from the Lie-Engel Theorem, since $X$ has to be solvable, and the reality of its roots in the $\Lambda^k$-representation forces the reality of its roots in the $\Lambda^1$-representation.