Proof relating to Orthogonal Complement 
a.) Show that if $A=A^T$ is a symmetric matrix, then $A\mathbf{x}=\mathbf{b}$ has a
  solution iff b is orthogonal to $\ker A$.
b.) Prove that if $K$ is a positive semi-definite matrix and $\mathbf{f}\notin
 \operatorname{rng}K$, then the quadratic function $$p(\mathbf{x}) = \mathbf{x}^\mathrm{T}K\mathbf{x} -2\mathbf{x}^\mathrm{T}\mathbf{f} + c$$ has no
  minimum value.
c.) Suppose $\{\mathbf{v_1},\ \cdots,\ \mathbf{v_n}\}$ span a subspace $V \subset \mathbb{R}^m$.
  Prove that $\mathbf{w}$ is orthogonal to $V$ iff $\mathbf{w}\in \operatorname{coker}A$ where
  $A=\begin{pmatrix}\mathbf{v_1} & \mathbf{v_2} & \cdots & \mathbf{v_n}\end{pmatrix}$ is the matrix with the indicated columns.

My attempt:
a.) We know that $A=A^\mathrm{T}$, then a vector $\mathbf{x}\in \mathbb{R}^n$ lies in $\ker A$ iff $A\mathbf{x} = \mathbf{0}$. 
By matrix multiplication we know that the $i^{\text{th}}$ entry of $A\mathbf{x}$ equals the vector product of the $i^{\text{th}}$ row $\mathbf{r_i}^T$ of $A$ and $\mathbf{x}$, hence $\mathbf{r_i}^{T}\cdot \mathbf{x} = \mathbf{r_i} \cdot  \mathbf{x} = 0$ iff $\mathbf{x}$ is orthogonal to $\mathbf{r_i}$. 
Therefore $\mathbf{x}\in \ker A$ iff $\mathbf{x}$ is orthogonal to all the rows of $A$.  Thus $A\mathbf{x} = \mathbf{b}$ has a solution iff $\mathbf{b}$ is orthogonal to $\ker A$. Is this correct?
b.) I do not know how to do this
c.) In this do I have to prove that $\mathbf{w} \in \operatorname{coker}A$ is orthogonal to the range of $A$? I am not exactly sure what they are asking. 

I'll assume $A$ has real entries.
For part a),
\begin{align*}
& u \in R(A)^{\perp} \\
\iff & \langle u, y \rangle = 0 \, \forall \, y \in R(A) \\
\iff & \langle u, Ax \rangle = 0 \, \forall \, x \\
\iff & \langle A^T u, x \rangle = 0 \, \forall \, x \\
\iff & A^T u = 0 \\
\iff & u \in N(A^T).
\end{align*}
This shows that $N(A^T)$ is the orthogonal complement of $R(A)$.  Because $A$
is symmetric, $N(A)$ is the orthogonal complement of $R(A)$.  Hence $b \in R(A)$
if and only if $b$ is orthogonal to $N(A)$.
For part b), let $v$ be the projection of $f$ onto $N(A)$.  Note that $v^T f = \|v\|^2 \neq 0$
and the function
\begin{align*}
g(\alpha) &= p(\alpha v) \\
&= -2\alpha v^T f + c
\end{align*}
is unbounded below.
