What are the conditions on matrix A so that for every b, Ax=b has a solution

A) is obviously false. Assume you already have a matrix $A$ such there there is a solution $x$ of $Ax=b$ for each $b\in\mathbb{R}^m.$ Then you could "extend" the matrix to the right with copies of columns of itself and still easily find solutions of the new linear system of equations.

B) is false for the same reason.

C) is true. It is implied by D)

D) is true. Assume the rows were dependent. Then there would be $u\in\mathbb{R}^m,\;\;u\neq 0$ such that $u^TA=0.$ If we can find a solution for every $b\in\mathbb{R}^m,$ then we can find a solution for $b=u.$ So let us assume there was an $x$ that satisfies $Ax=u.$ We would have $0=0x=u^TAx=u^Tu\neq 0.$ This is a contradiction, the rows must be linear independent.

E) is true. It is just another way of saying that for each $b\in\mathbb{R}^m,$ there exists an $x\in\mathbb{R}^n$ such that $Ax=b.$

F) is false. The subspace of $\mathbb{R}^n$ spanned by the rows of $A$ is at most $m$-dimensional (as we have only $m$ rows). And we have already seen that we can have $n>m.$

For G) and H), I assume that "columns/rows are not zeroes" means "columns/rows do not consist of zeros, only"

G) is false. Similar to A). I can append arbitrarily many columns full of zeroes to a matrix $A$ that "works" and still find a solution for each $b.$

H) is true. If I had a row full of zeroes in $A$, then the result of $Ax$ would also be $0$ in the same row. This would limit the possibility to get all possible $b.$