covering group of a topological group $H$: (locally) path-connected, and semilocally simply connected when reduced to $\pi_0(H)=\pi_1(H)=0?$

Question

According to covering group on Wikipedia, we conern the covering group of a topological group.

If $H$ is a

1. path-connected,
2. locally path-connected, and
3. semilocally simply connected group,

then it has a universal cover. By the previous construction the universal cover can be made into a topological group with the covering map a continuous homomorphism. This group is called the universal covering group of $H$.

My question is that on the conditions:

1. "If H is a path-connected," in which case path-connected is equivalent to the homotopy group $$\pi_0(H)=0?$$

2. "If H is a locally path-connected," in which case locally path-connected is equivalent to the homotopy group $$\pi_0(H)=0?$$

Are there cases in the topological group, such that the path-connected, and/or locally path-connected, is not $\pi_0(H)=0?$

Are there cases in the topological group, such that the path-connected, and/or locally path-connected, is, if an only if, $\pi_0(H)=0?$

3. "If H is a semilocally simply connected" in which case semilocally simply connected is equivalent to the homotopy group $$\pi_1(H)=0?$$

Are there cases in the topological group, such that the semilocally simply connected, is not $\pi_1(H)=0?$

Are there cases in the topological group, such that the semilocally simply connected, is, if an only if, $\pi_1(H)=0?$