Largest weight-limited connected subgraph: NP-complete?
The decision version is NP-complete so the problem is NP-hard. The following is a reduction from 3SAT.

For an instance of 3SAT with $n$ variables $x_1,\ldots, x_n$ and $m$ clauses, create a graph as follows:


For each variable $x_i$ create two vertices $v_i^0, v_i^1$ with cost $1$. Create an edge between each two vertices of these $2n$ vertices, i.e. the $2n$ vertices induce a clique.
For each variable $x_i$ create a vertex $v_i$ with cost $0$, and two edges $(v_i, v_i^0)$ and $(v_i,v_i^1)$.
For each clause $c_l$ with variables $x_i,x_j,x_k$, create a vertex $u_l$ with cost $0$, and three edges $(u_l, v_i^{b_i})$ (if the literal is positive, $b_i=1$, otherwise $b_i=0$, and the same below), $(u_l, v_j^{b_j})$ and $(u_l, v_k^{b_k})$.


Then we ask if there is a connected vertex set $T$ with size at least $2n+m$ and the sum of cost is no more than $n$.

Now if the instance of 3SAT has a feasible solution where $x_i=b_i$, we can make $T$ consist of all $v_i$s, $u_l$s and $v_i^{b_i}$s.

On the other hand, if $T$ exists, because you can afford at most $n$, at most $n$ vertices among $v_i^0, v_i^1$s are chosen, thus all $n+m$ vertices with cost $0$ must be chosen. This means:


Since $v_i$ is chosen, at least one vertex of $v_i^0$ and $v_i^1$ is chosen. Because at most $n$ vertices among $v_i^0, v_i^1$s are chosen, exactly one of $v_i^0$ and $v_i^1$ is chosen. We assign $x_i$ accordingly.
Since $u_l$ is chosen, at least one corresponding literal has value $1$, thus the clause is satisfied.


So the instance of 3SAT has a feasible solution.