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Here are some exercises of algebraic topology (they aren't homework but self-studying)

a) Calculate the fondamental group of the subspace of $$\mathbb{R}^3 H = S^{2} \bigcup \{ \text{the 3 coordinate planes} \}$$

b) Prove that for every $ 0 \le n \le m - 3 \ $, the complementary in $\mathbb{R}^{m} \ $ of a $n$-dimensional vector subspace is simply connected.

I'm particularly interested in the second point. Any hint ?

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For (a) If you collapse the $3$ coordinate planes to a point, the space you are left with is homeomorphic to the wedge of $8$ spheres, which is simply connected. Thus $\pi_1(H)$ is trivial.

For (b) Let $\mathbb{R}^n\subseteq \mathbb{R}^m$ be the collection of points $(x_1,\ldots, x_m)$ such that $x_{n+1} = \cdots = x_m = 0$. I claim that there is a deformation retraction of $\mathbb{R}^m\smallsetminus \mathbb{R}^n$ onto the subset $\{(x_1,\ldots, x_m) : x_1 = \cdots = x_n = 0\}\smallsetminus\{0\}\cong \mathbb{R}^{m-n}\smallsetminus \{0\}$. This deformation retraction is given by $(x_1,\ldots, x_m)\mapsto (tx_1,\ldots, tx_n, x_{n+1},\ldots, x_m)$ with $t\in [0,1]$. It follows that $\pi_1(\mathbb{R}^m\smallsetminus \mathbb{R}^n)\cong \pi_1(\mathbb{R}^{m-n}\smallsetminus \{0\})$. Now $\mathbb{R}^{m-n}\smallsetminus\{0\}$ is homotopy equivalent to $\mathbb{S}^{m-n-1}$, which is simply connected so long as $m-n-1>1$, i.e., so long as $m\geq n + 3$. Therefore $\pi_1(\mathbb{R}^{m}\smallsetminus \mathbb{R}^n)$ is trivial when $m\geq n+3$.

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