Prove: It exist $ n\geq0$ and $ \lambda_0,\dots,\lambda_n\in\mathbb{Q}$, so that $ \varphi = \sum_{i=0}^n\lambda_i(x^i)^*$.

Question

Could you guys help me with this task?

Let $ \mathbb{Q}[x]$ be the vector space of the polynomials and let $ \varphi:\mathbb{Q}[x]\to\mathbb{Q}$ be the linear function given by $ \varphi(P) = P(2)$. Consider the base $ B = \{ x^i: i\geq0\}$ of $ \mathbb{Q}[x]$. $\forall$ $ i\geq0$ we denote with $ (x^i)^*$ the linear function given by $ (x^i)^*(x^i) = 1$ and $ (x^i)^*(x^j) = 0$ $\forall$ $ j\neq i$.

Decide which of the statement is true or false:

1. It exist $ n\geq0$ and $ \lambda_0,\dots,\lambda_n\in\mathbb{Q}$, so that $ \varphi = \sum_{i=0}^n\lambda_i(x^i)^*$.
2. $\forall $ $ n\geq0$ ist the set $ \{(x^i)^* : 0\leq i\leq n\}$ a Basis of $ \{ P \in \mathbb{Q}[x]: P=0$ or degree$ (P)\leq n\}^*$.

Thanks in advance!