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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.

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 just '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 (because $\Lambda^k(X)$ is and it is either isomorphic to $X$ or, if $k$ is even and $-I$ lies in $X$, then $X$ is a double cover of $\Lambda^k(X)$), and the reality of its roots in the $\Lambda^k$-representation forces the reality of its roots in the $\Lambda^1$-representation.

As far as Question 2 goes, it's not at all clear what the best estimate would be, but it's clearly better than the trivial bound. To get any meaningful bound, you'd need to know something about the reductive group $G$. For example, if $G$ is finite, then there are only a finite number of irreducible (faithful) representations of $G$ and so there's only a finite list of things to consider. in that case, 'asymptotic bounds' won't make much sense. If $G$ is, say, a simple Lie group, such as $\mathrm{SL}(2,\mathbb{R})$, then the number of irreducible components of $\Lambda^k(V)$ could only grow polynomially (maybe even only quadratically, but I'm not sure about that) with the dimension of $V$, so the binomial coefficients estimate is pretty far off. Probably, for higher rank groups, there is an even lower estimate that can be got via multiplicity formulae, but I don't know whether it's qualitatively better than crude estimates based on primary $sl(2)$-subgroups.

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 $J^2=-1$. Then $J$ is not conjugate to any upper triangular matrix (over $\mathbb{R}$) since it does not have real eigenvalues. Meanwhile, 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.

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.

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 $J^2=-1$. Then $J$ is not conjugate to any upper triangular matrix (over $\mathbb{R}$) since it does not have real eigenvalues. Meanwhile, the eigenvalues of $\Lambda^2(J):\Lambda^2(\mathbb{R}^4)\to\Lambda^2(\mathbb{R}^4)$ are $\pm 1$, and $\Lambda^2(J)$ has a basis of eigenvectors.

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.

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 $J^2=-1$. Then $J$ is not conjugate to any upper triangular matrix (over $\mathbb{R}$) since it does not have real eigenvalues. Meanwhile, 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.