Sorry for the lack of latex. The question I want to ask would need all this info and it would take very long to write it.
(a) Irrotational flow means $\nabla \times \textbf u =0$ so we can define potential to be $\textbf u =\nabla \phi$ and since it is irrotational and inviscid, the continuity equation holds so $\nabla \cdot \textbf u =0$ so the laplacian of the potential holds, $\nabla \cdot \textbf u =\nabla \cdot \nabla \phi=\nabla^2 \phi=0$.
(b) $u_r = U\cos \theta (1-a^3 / r^3)$ and $u_{\theta}=-U\sin \theta (1+a^3/2r^3)$
(c) $(r=a, \theta = 0)$ and $(r=a, \theta = \pi)$
(d) $ψ=\frac U2 \sin^2 \theta (r^2 - a^3/r)$
(e) Would it be something like this: 
I am mainly stuck on c, d, and e. I am a tiny bit unsure on part (a) but it seems to make sense. Please do say if not.
Please help.
For (b), we can take partial derivatives of $\phi$ to obtain
$$u_r = \frac{\partial \phi}{\partial r} = U \cos \theta - U\frac{a^3}{r^3} \cos \theta, \\ u_\theta = \frac{1}{r} \frac{\partial \phi}{\partial \theta}= -U \sin \theta - U\frac{a^3}{2r^3} \sin \theta.$$
For (c), on the surface of the sphere $r = a$ and the radial velocity component vanishes, $u_r(a, \theta) = 0,$ since this component is normal to the surface. This is apparent upon substituting $a$ for $r$ in the above expression for $u_r$. The tangential component is
$$u_\theta(a, \theta) = -\frac{3}{2}U \sin \theta.$$
We have a stagnation points on the surface $r = a$ where both velocity components vanish: $\theta = 0$ (fore) and $\theta = \pi$ (aft).
For (d), we solve for the streamfunction using
$$\frac{\partial \psi}{\partial \theta} = r^2 \sin \theta u_r = U\sin \theta \cos \theta \left(r^2 - \frac{a^3}{r} \right) = \frac{U}{2} \sin 2 \theta\left(r^2 - \frac{a^3}{r} \right)\\ \implies \psi = -\frac{U}{4} \cos 2 \theta\left(r^2 - \frac{a^3}{r} \right) + F(r) \\ \implies \psi = -\frac{U}{4} (1 - 2\sin^ 2 \theta)\left(r^2 - \frac{a^3}{r} \right) + F(r) \\ \implies \psi = \frac{U}{2} \sin^ 2 \theta\left(r^2 - \frac{a^3}{r} \right) - \frac{U}{4}\left(r^2 - \frac{a^3}{r} \right) + F(r)$$
and
$$\frac{\partial \psi}{\partial r} = -r \sin \theta u_\theta = U\sin^2 \theta\left(r + \frac{a^3}{2r^2} \right) \\ \implies \psi = U \sin^2 \theta \left(\frac{r^2}{2} - \frac{a^3}{2r} \right) + G(\theta),$$
We get a consistent solution by choosing
$$G(\theta) = 0, \\ F(r) = \frac{U}{4}\left(r^2 - \frac{a^3}{r} \right),$$
and it follows that
$$\psi = \frac{U}{2} \sin^2 \theta \left(r^2 - \frac{a^3}{r} \right)$$