1. One can find a cone in $\R^3$ that is tangent to the sphere at the curve, or do the computation using the pullback metric tensor. The following might be helpful:

1. if $A$ is square matrix, to compute $\exp(A)$, we better do eigenvalue decomposition $A = U D U^{-1}$ where $D$ is diagonal,then $e^A = U e^D U^{-1}$ (proof by Taylor expansion).
2. It is useful to use an orthonormal frame for the tangent bundle, i.e, Cartan's moving frame. Then the coefficients changes in a nice way.

2 and 3: Cartan's moving frame is useful. Since in two-dimension, we have $\omega_1^2 = -\omega_2^1$ only one component to solve. Then, one can compute the curvature $R = d \omega + \omega \wedge \omega.$

4. This one is a bit more difficult. Let $X(t) = (L_{\exp(t Y / 2)})_* (R_{\exp(t Y / 2)})_* X$.

1. To verify the equation at time $t$, we need to relate $X(t+\epsilon)$ and $X(t)$.
2. Recall that conjugation action is $Conj_g = L_g \circ R_{g^{-1}} = R_{g^{-1}} \circ L_g$
3. $X(t+\epsilon) = (L_{\exp(\epsilon Y / 2)})_* (R_{\exp(\epsilon Y / 2)})_* X(t) = (Conj_{\exp(-\epsilon Y / 2)})_* (L_{\exp(\epsilon Y )})_* X(t)$
4. Try to use left-invariant vector fields to trivialize $TG$, and write down the parallel transport equations.