Answer
$$\cos \theta=\frac{\sqrt 5}{3}.$$
Work Step by Step
Let $f(x)=1, \quad g(x)=x^2$, then
$$\langle f, g\rangle=\int_{-1}^{1}f(x)g(x) d x=\int_{-1}^{1} x^{2}d x=\left[\frac{1}{3}x^{3}\right]_{-1}^{1}=\frac{2}{3}$$
and
$$\langle f, f\rangle=\int_{-1}^{1} 1 d x=\left.x\right|_{-1} ^{1}=2 \Longrightarrow\| f \| =\sqrt{\langle f, f\rangle}=\sqrt 2$$
$$\langle g, g\rangle=\int_{-1}^{1} x^4 d x=\left.\frac{1}{5}x^5\right|_{-1} ^{1}=\frac{2}{5}\Longrightarrow\| g \| =\sqrt{\langle g, g\rangle}=\sqrt{\frac{2}{5}}$$
The angle $\theta$ between $f$ and $g$ is given by the formula
$$\cos \theta=\frac{\langle f, g\rangle}{\|f\| \cdot\|g\|}=\frac{\frac{2}{3}}{\sqrt{2}\sqrt{\frac{2}{5}} }=\frac{\sqrt 5}{3}.$$