Let $V,W$ be complex inner product spaces. Fix $v \in V$ and $w \in W$ and define $T:V \rightarrow W$ by $T(u)=\langle u,v\rangle_Vw$. Find $T^*:W \rightarrow V$.
By the properties of $T^*$, we must have that $\langle T(a),b \rangle=\langle a,T^*(b) \rangle$, so therefore, $\langle T(a),b \rangle = \langle \langle a,v \rangle_Vw,b \rangle$.
This is where I get stuck because I can't figure out how to reduce this or manipulate it so I can figure out what $T^*$ is. So my question is, how do I continue my reasoning to find $T^*$?
You need to use the fact that your complex inner products are linear in the first argument and conjugate-linear in the second argument. Using linearity of $\langle \cdot,\cdot \rangle_W$ in the first argument, you get $$ \langle T(a),b \rangle_W = \langle \langle a,v \rangle_V w,b \rangle_W = \langle a,v \rangle_V \langle w, b \rangle_W. $$ Can you now use conjugate-linearity of $\langle \cdot,\cdot \rangle_V$ in the second argument to rewrite $\langle a,v \rangle_V \langle w, b \rangle_W$ as $\langle a, x \rangle_V$ for some $x \in V$? For then, by definition, $T^\ast(b) = x$, whatever $x$ may be.