The following steps lead to a solution:
(1) Prove that the tuple $(1,x,x^2,\dots)$ is a basis for the complex vector space $V$ of all polynomials (in the variable $x$) with coefficients in $\mathbb{C}$.
(2) If $p(x)=\sum_{i=0}^n a_ix^i$ and if $\Lambda$ is a linear functional on $V$, prove that $\Lambda(p(x))=\sum_{i=0}^n a_i\Lambda(x^i)$. (Hint: the linearity of the functional $\Lambda$ is, of course, relevant.)
(3) Therefore, the linear functional $\Lambda$ is completely determined by the values $\Lambda(x^i)$ for $i\geq 0$ an integer.
The following exercises are relevant:
Exercise 1: Prove the Riesz representation theorem for finite dimensional Hilbert spaces:
If $H$ is a finite dimensional Hilbert space and if $\Lambda$ is a linear functional on $H$, then there exists a unique vector $y\in H$ such that $\Lambda(x)=\langle x,y \rangle$ for all $x\in H$. (Note that $\langle \cdot,\cdot \rangle$ is the inner product on $H$.)
(Hint: note that $H$ has an orthonormal basis.)
Exercise 2: Let $V$ be a finite dimensional vector space over $\mathbb{C}$ and let $(v_1,\dots,v_n)$ be a basis of $V$. Prove that if $(a_1,\dots,a_n)$ is a tuple of complex numbers, then the map $\Lambda:V\to\mathbb{C}$ given by $\Lambda(\sum_{i=1} c_iv_i)=\sum_{i=1}^n c_ia_i$ (where $c_i\in\mathbb{C}$ for all $1\leq i\leq n$) is a linear functional on $V$. Conversely, prove that all linear functionals on $V$ have this form.
Exercise 3: Let $V$ be a finite dimensional vector space over $\mathbb{C}$ and let $(v_1,\dots,v_n)$ be a basis of $V$. If $1\leq i\leq n$, let $\Lambda_i:V\to \mathbb{C}$ be defined by the rule $\Lambda_i(v_i)=1$ and $\Lambda_i(v_j)=0$ if $j\neq i$. Prove that $\Lambda_i$ is a linear functional on $V$ for all $1\leq i\leq n$. If $V^{*}$ is the dual space of $V$ (= vector space of all linear functionals on $V$), prove that $(\Lambda_1,\dots,\Lambda_n)$ is a basis of $V^{*}$.
Exercise 4: Let $V$ be the vector space of all polynomials (in the variable $x$) with coefficients in $\mathbb{C}$. If $i\geq 0$, let $\Lambda_i:V\to \mathbb{C}$ be defined by the rule $\Lambda_i(x^i)=1$ and $\Lambda_i(x^j)=0$ if $j\neq i$. Prove that $\Lambda_i$ is a linear functional on $V$ for all $i\geq 0$. Is the tuple $(\Lambda_0,\Lambda_1,\Lambda_2,\dots)$ a basis of the dual space of $V$?
I hope this helps!
You have the right ideas. Indeed, your claim is not true. Consider, for example, the transformation
$$
T = \pmatrix{0&1\\0&0}
$$
Verify that im$(T) = \ker(T)$, and that both of these are one-dimensional subspaces of $\Bbb R^2$.
Notably, however, the statement will hold for any self-adjoint (symmetric) operator $T:V \to V$.
Best Answer
A counterexample to (ii):
Consider a $2$-dimensional vector space $V$ and distinct $1$-dimensionsal subspaces $A, B,C$. The $B+C=V$, so $A\cap(B+C)=A$, but $A\cap B, \:A\cap C=\{0\}$, so that $A\cap B+A\cap C=\{0\}$.
Of course, this is also a counterexample to (i).