Okay, this got a bit mangled.
(1) Is the set $\mathbf{V}=\{A\mid A\text{ is an }n\times n\text{ matrix and }\det(A)=0\}$ a vector space, under the usual addition and scalar multiplication of matrices?
Since the set of all $n\times n$ matrices is a vector space, the question is really whether this is a subspace (all the axioms of a vector space will necessarily hold, except perhaps for the existence of a zero vector, the existence of inverses, and the "hidden" axioms that the set must be closed under vector addition and scalar multiplication: the sum of two vectors in $\mathbf{V}$ must lie in $\mathbf{V}$, and every scalar multiple of a vector in $\mathbf{V}$ lies in \mathbf{V}).
Scalar multiplication is easy: if \alpha\in\mathbb{R}$ and $A$ is any $n\times n$ matrix, then we know that $\det(\alpha A) = \alpha^n \det(A)$. So $\det(\alpha A)=0$ if and only if $\alpha = 0$ or $\det(A)=0$. So, if $A\in\mathbf{V}$, then $\alpha A\in\mathbf{V}$ for all scalars $\alpha.
What about vector addition? We would need to show that if A$ and $B$ both have nonzero determinant, then so does $A+B$. But this is not the case, as Agustí Roig points out. It should be easy to come up with a similar example for any $n\gt 0$. So $\mathbf{V}$ is not closed under addition (exhibit an explicit pair of matrices, both in $\mathbf{V}$, but whose sum is not in $\mathbf{V}$; sometimes the sum of two matrices in $\mathbf{V}$ is in $\mathbf{V}$, the point is that it doesn't always lie in $\mathbf{V}, so give an example!).
(2). Again, since all real valued functions form a vector space, the only issue is whether the set \mathbf{V} = \{ f(x)\mid f(x) = (ax+b)e^{-x},\ a,b\in\mathbb{R}\} is closed under sums and scalar multiplication.
Suppose f(x),g(x)\in\mathbf{V}$. Will $f(x)+g(x)$ lie in $\mathbf{V}$ as well? Write \begin{align*} f(x) &= (ax+b)e^{-x}\\ g(x) &= (cx+d)e^{-x}\\ f(x)+g(x) &= \Bigl( (ax+b)e^{-x}\Bigr) + \Bigl( (cx+d)e^{-x}\Bigr)\\ &= \Bigl( (ax+b) + (cx+d)\Bigr) e^{-x}\\ &= \Bigl( (a+c)x + (b+d)\Bigr) e^{-x}. \end{align*} So setting $A=a+c$ and $B=b+d$, which are real numbers because all of $a,b,c,d$ are real numbers, then we see that we can write $f(x)+g(x)$ in the form $(Ax+B)e^{-x}$. So if each of $f(x)$ and $g(x)$ are in $\mathbf{V}$, then $f(x)+g(x)\in\mathbf{V}.
Now you need to show that if f(x)\in\mathbf{V}$ and $\alpha\in\mathbb{R}$, then $\alpha f(x)\in\mathbf{V}. I'll leave that to you to do.
What about the dimension of \mathbf{V}$? If you want to describe an element of $\mathbf{V}$, you really only need to specify two things: the value of $a$ and the value of $b$. that suggests that the dimension will be $2$. Can you find a set of $2$ linearly independent functions, both in $\mathbf{V}$, that span $\mathbf{V}$?